python刷题+leetcode(第三部分)

200.最大正方形

思路:与岛屿，水塘不同的是这个相对要规则得多，而不是求连通域，所以动态规划构造出状态转移方程即可

动态规划　if 0, dp[i][j] =0

if 1, dp[i][j] = min(dp[i-1][j-1],dp[i-1][j],dp[i][j-1])+1

class Solution:
    def maximalSquare(self, matrix):
        print('==np.array(matrix):\n', np.array(matrix))
        h = len(matrix)
        w = len(matrix[0])
        max_side = 0
        dp = [[0 for j in range(w)] for i in range(h)]
        print('==dp:', np.array(dp))

        for i in range(h):
            for j in range(w):
                if matrix[i][j] == '1' and i>0 and j>0:
                    dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1
                    max_side = max(max_side, dp[i][j])
                elif i==0:
                    dp[i][j] = int(matrix[i][j])
                    max_side = max(max_side, dp[i][j])
                elif j==0:
                    dp[i][j] = int(matrix[i][j])
                    max_side = max(max_side, dp[i][j])
                else:
                    pass
        print('==dp:', np.array(dp))
        # print(max_side)
        return max_side**2


matrix = [["1", "0", "1", "0", "0"], ["1", "0", "1", "1", "1"], ["1", "1", "1", "1", "1"], ["1", "0", "0", "1", "0"]]

sol = Solution()
sol.maximalSquare(matrix)

201.统计全为 1 的正方形子矩阵

思路:动态规划　if 0, dp[i][j] =0

if 1, dp[i][j] = min(dp[i-1][j-1],dp[i-1][j],dp[i][j-1])+1

为１的时候　res自加1,再加上dp[i][j]增加的部分，也可以看成是dp的和

class Solution:
    def countSquares(self, matrix):
        print('==np.array(matrix)\n', np.array(matrix))
        h = len(matrix)
        w = len(matrix[0])
        dp = [[0 for i in range(w)] for j in range(h)]
        print('==np.array(dp):', np.array(dp))
        res = 0
        for i in range(h):
            for j in range(w):
                if matrix[i][j] == 1 and i > 0 and j > 0:
                    res += 1
                    dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1
                    res += dp[i][j] - 1  # 减去1代表增加的矩形个数
                elif i==0:
                    dp[i][j] = matrix[i][j]
                    if matrix[i][j] == 1:
                        res+=1
                elif j==0:
                    dp[i][j] = matrix[i][j]
                    if matrix[i][j] == 1:
                        res+=1
                else:
                    pass
        print('==np.array(dp):', np.array(dp))
        print('==after res:', res)
        return res


matrix = [
    [0, 1, 1, 1],
    [1, 1, 1, 1],
    [0, 1, 1, 1]
]
sol = Solution()
sol.countSquares(matrix)

203.分发饼干

思路:排序加贪心　先给胃口小的饼干

#思路:排序加贪心 先让胃口小的孩子满足
class Solution:
    def findContentChildren(self, g, s):
        print('==g:', g)
        print('==s:', s)
        g = sorted(g)#孩子
        s = sorted(s)#饼干
        res = 0
        for j in range(len(s)):#遍历饼干　先给胃口小的分配
            if res<len(g):
                if g[res]<=s[j]:
                    res+=1
        print('==res:', res)
        return res

    
g = [1,2]
s = [1,2,3]
# g = [1, 2, 3]
# s = [1, 1]
sol = Solution()
sol.findContentChildren(g, s)

205.同构字符串

思路:hash　　构造映射关系

class Solution:
    def isIsomorphic(self, s: str, t: str) -> bool:
        if len(s) != len(t):
            return False
        dic = {}
        for i in range(len(s)):
            if s[i] not in dic:#未出现过
                if t[i] in dic.values():#value已经出现过之前构造的dict中了
                    return False
                dic[s[i]] = t[i]
                
            else:#出现过
                 if dic[s[i]]!=t[i]:
                     return False
        return True

# s = "egg"
# t = "add"
s="ab"
t="aa"

sol = Solution()
sol.isIsomorphic(s, t)

 

208.最后一块石头的重量

思路1:while循环　排序从大到小　一直取前两块石头进行比较

class Solution:
    def lastStoneWeight(self, stones):
        while len(stones)>=2:
            stones = sorted(stones)[::-1]
            if stones[0]==stones[1]:
                stones=stones[2:]
            else:
                stones = [stones[0]-stones[1]]+stones[2:]
        print('===stones:', stones)
        return stones[-1] if len(stones) else 0



stones = [2, 7, 4, 1, 8, 1]
# stones = [2, 2]
sol = Solution()
res= sol.lastStoneWeight(stones)
print('==res:', res)

更简洁写法：

class Solution:
    def lastStoneWeight(self, stones: List[int]) -> int:
        while len(stones) >= 2:
            stones = sorted(stones)
            stones.append(stones.pop() - stones.pop())
        return stones[0]

思路2:堆队列，也称为优先队列

import heapq
class Solution:
    def lastStoneWeight(self, stones):
        h = [-stone for stone in stones]
        heapq.heapify(h)

        print('==h:', h)
        while len(h) > 1:
            a, b = heapq.heappop(h), heapq.heappop(h)
            if a != b:
                heapq.heappush(h, a - b)
            print('==h:', h)
        return -h[0] if h else 0



stones = [2, 7, 4, 1, 8, 1]
# stones = [2, 2]
sol = Solution()
res= sol.lastStoneWeight(stones)
print('==res:', res)

210.无重叠区间

class Solution:
    def eraseOverlapIntervals(self, intervals):

        n = len(intervals)
        if n<=0:
            return 0
        intervals = sorted(intervals, key=lambda x: x[-1])
        print('==intervals:', intervals)
        
        res = [intervals[0]]
        for i in range(1, n):
            if intervals[i][0] >= res[-1][-1]:
                res.append(intervals[i])
        print(res)
        return n - len(res)


# intervals = [[1, 2], [2, 3], [3, 4], [1, 3]]
# intervals = [[1,100],[11,22],[1,11],[2,12]]
intervals = [[0, 2], [1, 3], [2, 4], [3, 5], [4, 6]]
sol = Solution()
sol.eraseOverlapIntervals(intervals)

212.种花问题

思路:判断是否是1或0,1就一种情况,0有两种情况

100 先判断1

 01000,要判断i为0时,i+1是否为1,否则说明就是001这种情况

# 100 先判断1
# 01000,要判断i为0时,i+1是否为1,否则说明就是001这种情况
class Solution:
    def canPlaceFlowers(self, flowerbed, n):

        i = 0
        res = 0
        while i<len(flowerbed):
            if flowerbed[i]==1:
                i+=2
            else:
                if i+1<len(flowerbed) and flowerbed[i+1]==1:
                    i+=3
                else:
                    res+=1
                    i+=2

        return True if res>=n else False

# flowerbed = [1,0,0,0,1]
# n = 1

# flowerbed = [1,0,0,0,1]
# n = 2

# flowerbed = [1,0,0,0,0,0,1]
# n = 2
flowerbed = [1,0,0,0,1,0,0]
n = 2
sol = Solution()
res= sol.canPlaceFlowers(flowerbed, n)
print('==res:', res)

216.较大分组的位置

其实就是再求聚类

思路1:动态规划

class Solution:
    def largeGroupPositions(self, s):
        dp = [0] * len(s)
        for i in range(1, len(s)):
            if s[i] == s[i - 1]:
                dp[i] = 1

        dp.append(0)
        print('==dp:', dp)
        index = [j for j in range(len(dp)) if dp[j] == 0]
        print('index:', index)
        res = []
        for k in range(len(index) - 1):
            if index[k + 1] - index[k] >= 3:
                res.append([index[k], index[k + 1] - 1])
        print('=res:', res)
        return res


s = "abbxxxxzzy"
sol = Solution()
sol.largeGroupPositions(s)

思路2:双指针

# 双指针
class Solution:
    def largeGroupPositions(self, s):
        res = []
        left, right = 0, 0
        while left < len(s):
            right = left + 1
            while right < len(s) and s[right] == s[left]:
                right += 1
            if right - left >= 3:
                res.append([left, right - 1])
            # 左指针跑到右指针位置
            left = right
        print('==res:', res)
        return res

s = "abbxxxxzzy"
# s = "abcdddeeeeaabbbcd"
sol = Solution()
sol.largeGroupPositions(s)

217.省份数量

思路：

可以把 n 个城市和它们之间的相连关系看成图， #

城市是图中的节点，相连关系是图中的边，

 给定的矩阵isConnected 即为图的邻接矩阵，省份即为图中的连通分量。

利用dfs将一个数组view遍历过的城市置位1。

# dfs
# 可以把 nn 个城市和它们之间的相连关系看成图，
# 城市是图中的节点，相连关系是图中的边，
# 给定的矩阵isConnected 即为图的邻接矩阵，省份即为图中的连通分量。
class Solution:
    def travel(self, isConnected, i, n):
        self.view[i] = 1  # 表示已经遍历过
        for j in range(n):
            if isConnected[i][j] == 1 and not self.view[j]:
                self.travel(isConnected, j, n)

    def findCircleNum(self, isConnected):
        n = len(isConnected)
        self.view = [0] * n
        res = 0
        for i in range(n):
            if self.view[i] != 1:
                res += 1
                self.travel(isConnected, i, n)
        print('==res:', res)
        return res

# isConnected = [[1, 1, 0],
#                [1, 1, 0],
#                [0, 0, 1]]
isConnected = [[1,0,0],
               [0,1,0],
               [0,0,1]]
sol = Solution()
sol.findCircleNum(isConnected)

218.旋转数组

思路1:截断拼接，注意的是一些边界条件需要返回原数组

class Solution:

    def rotate(self, nums: List[int], k: int) -> None:
        if len(nums)<=1 or k==0 or k%len(nums)==0:
            return nums
        n = len(nums)
        k = k%n
        # print(nums[-k:]+nums[:n-k])
        nums[:] = nums[-k:]+nums[:n-k]
        return nums

思路2:先左翻转，在右翻转，在整体翻转 

class Solution:

    def reverse(self, i, j, nums):#交换位置的
        while i < j:#
            nums[i], nums[j] = nums[j], nums[i]
            i += 1
            j -= 1

    def rotate(self, nums, k):
        """
        Do not return anything, modify nums in-place instead.
        """
        n = len(nums)
        k %= n #有大于n的数
        self.reverse(0, n - k - 1, nums) #左翻
        self.reverse(n - k, n - 1, nums) #右翻
        self.reverse(0, n - 1, nums) #整体翻
        print(nums)
        return nums

# nums = [1,2,3,4,5,6,7]
# k = 3
nums = [1,2,3,4,5,6]
k = 11
sol = Solution()
sol.rotate(nums, k)

219．汇总区间

思路:双指针

class Solution:
    def summaryRanges(self, nums):
        res = []
        left =0
        right = 0
        while right<len(nums):
            right = left+1
            while right<len(nums) and nums[right] - nums[right-1] == 1:
                right+=1
            if right -1>left:
                res.append(str(nums[left]) + "->" + str(nums[right-1]))
            else:
                res.append(str(nums[left]))
            left = right
        print(res)
        return res
nums = [0,1,2,4,5,7]
sol = Solution()
sol.summaryRanges(nums)

220.冗余连接

思路:并查集

#并查集:合并公共节点的,对相邻节点不是公共祖先的进行合并
class Solution:
    def find(self, index):  # 查询
        if self.parent[index] == index:  # 相等就返回
            return index
        else:
            return self.find(self.parent[index])  # 一直递归找到节点index的祖先

    def union(self, i, j):  # 合并
        self.parent[self.find(i)] = self.find(j)

    def findRedundantConnection(self, edges):
        nodesCount = len(edges)
        self.parent = list(range(nodesCount + 1))
        print('==self.parent:', self.parent)

        for node1, node2 in edges:
            print('==node1, node2:', node1, node2)
            if self.find(node1) != self.find(node2):#相邻的节点公共祖先不一样就进行合并
                print('===hahhaha===')
                self.union(node1, node2)
                print('=self.parent:', self.parent)
            else:
                return [node1, node2]

        return []


edges = [[1, 2], [1, 3], [2, 3]]
sol = Solution()
res = sol.findRedundantConnection(edges)
print('=res:', res)

223.可被 5 整除的二进制前缀

思路:二进制移位　在和5求余

class Solution:
    def prefixesDivBy5(self, A: List[int]) -> List[bool]:
        res = [False]*len(A)
        value = 0
        for i in range(len(A)):
            value = (value<<1) + A[i]
            # print(value)
            if value%5==0:
                res[i]=True
        return res

225.移除最多的同行或同列石头

思路1: 其实主要是算连通域的个数,当满足同行或者同列就算联通，
 输出的结果就是石头个数减去连通域个数,第一种解法超时

# 其实主要是算连通域的个数,当满足同行或者同列就算联通，
# 输出的结果就是石头个数减去连通域个数
#第一种直接dfs会超时
import numpy as np
class Solution:
    def dfs(self, rect, i, j, h, w):
        if i < 0 or i >= h or j < 0 or j >= w or rect[i][j] != 1:
            return
        rect[i][j] = -1
        for i_ in range(h):
            self.dfs(rect, i_, j, h, w)
        for j_ in range(w):
            self.dfs(rect, i, j_, h, w)

    def removeStones(self, stones):
        n = 10
        rect = [[0 for _ in range(n)] for _ in range(n)]
        print(len(rect))
        for stone in stones:
            rect[stone[0]][stone[-1]] = 1
        print('before np.array(rect):', np.array(rect))
        h, w = n, n
        graphs = 0
        for i in range(h):
            for j in range(w):
                if rect[i][j] == 1:
                    graphs += 1
                    self.dfs(rect, i, j, h, w)
        print('after np.array(rect):', np.array(rect))
        print(graphs)
        return len(stones) - graphs


stones = [[0, 0], [0, 1], [1, 0], [1, 2], [2, 1], [2, 2]]
sol = Solution()
res = sol.removeStones(stones)
print('===res:', res)

思路2:并查集

class Solution:
    # 并查集查找
    def find(self, x):
        if self.parent[x] == x:
            return x
        else:
            return self.find(self.parent[x])
    #合并
    def union(self,i, j):
        self.parent[self.find(i)] = self.find(j)
    def removeStones(self, stones):
        # 因为x,y所属区间为[0,10^4]
        # n = 10001
        n = 10
        self.parent = list(range(2 * n))
        for i, j in stones:
            self.union(i, j + n)
            print('==self.parent:', self.parent)

        # 获取连通区域的根节点
        res = []
        for i, j in stones:
            res.append(self.find(i))
        print('=res:', res)
        return len(stones) - len(set(res))


stones = [[0, 0], [0, 1], [1, 0], [1, 2], [2, 1], [2, 2]]
sol = Solution()
res = sol.removeStones(stones)
print('===res:', res)

226..缀点成线

思路:判断斜率 将除换成加

class Solution:
    def checkStraightLine(self, coordinates):
        n = len(coordinates)
        for i in range(1, n-1):
            if (coordinates[i+1][1]-coordinates[i][1])*(coordinates[i][0]-coordinates[i-1][0])=(coordinates[i][1]-coordinates[i-1][1])*(coordinates[i+1][0]-coordinates[i][0]):
                return False
        return True


coordinates = [[1,2],[2,3],[3,4],[4,5],[5,6],[6,7]]
sol = Solution()
sol.checkStraightLine(coordinates)

227.账户合并

思路:并查集

#思想是搜索每一行的每一个邮箱，如果发现某一行的某个邮箱在之前行出现过，那么把该行的下标和之前行通过并查集来合并，
class Solution(object):
    def find(self, x):
        if x == self.parents[x]:
            return x
        else:
            return self.find(self.parents[x])
    def union(self,i, j):
        self.parents[self.find(i)] = self.find(j)
    def accountsMerge(self, accounts):
        # 用parents维护每一行的父亲节点
        # 如果parents[i] == i, 表示当前节点为根节点
        self.parents = [i for i in range(len(accounts))]
        print('==self.parents:', self.parents)
        dict_ = {}

        # 如果发现某一行的某个邮箱在之前行出现过，那么把该行的index和之前行合并（union）即可
        for i in range(len(accounts)):
            for email in accounts[i][1:]:
                if email in dict_:
                    self.union(dict_[email], i)
                else:
                    dict_[email] = i
        print('===self.parents:', self.parents)
        print('=== dict_:', dict_)
        import collections
        users = collections.defaultdict(set)
        print('==users:', users)
        res = []
        # 1. users：表示每个并查集根节点的行有哪些邮箱
        # 2. 使用set：避免重复元素
        # 3. 使用defaultdict(set)：不用对每个没有出现过的根节点在字典里面做初始化
        for i in range(len(accounts)):
            for account in accounts[i][1:]:
                users[self.find(i)].add(account)
        print('==users:', users)
        # 输出结果的时候注意结果需按照字母顺序排序（虽然题目好像没有说）
        for key, val in users.items():
            res.append([accounts[key][0]] + sorted(list(val)))

        return res


accounts = [["John", "[email protected]", "[email protected]"],
            ["John", "[email protected]"],
            ["John", "[email protected]", "[email protected]"],
            ["Mary", "[email protected]"]]

# [["John", '[email protected]', '[email protected]', '[email protected]'],
#  ["John", "[email protected]"],
#  ["Mary", "[email protected]"]]

sol = Solution()
res= sol.accountsMerge(accounts)
print(res)

228.连接所有点的最小费用

思路1: 其实就是求最小生成树，首先想到的是kruskal 但是时间复杂度较高,超时

# 其实就是求最小生成树:采用kruskal 但是时间复杂度较高,超时
class Solution:
    def minCostConnectPoints(self, points):
        edge_list = []
        nodes = len(points)
        for i in range(nodes):
            for j in range(i):
                dis = abs(points[i][0] - points[j][0]) + abs(points[i][1] - points[j][1])
                edge_list.append([i, j, dis])
        print('==edge_list:', edge_list)
        edge_list = sorted(edge_list, key=lambda x: x[-1])
        print('==edge_list:', edge_list)
        group = [[i] for i in range(nodes)]
        print('==group:', group)
        res = 0
        for edge in edge_list:
            for i in range(len(group)):
                if edge[0] in group[i]:
                    m = i  # 开始节点
                if edge[1] in group[i]:
                    n = i  # 结束节点
            if m != n:
                # res.append(edge)
                res += edge[-1]
                group[m] = group[m] + group[n]
                group[n] = []
            print(group)
        print('==res:', res)
        return res

points = [[0, 0], [2, 2], [3, 10], [5, 2], [7, 0]]
sol = Solution()
sol.minCostConnectPoints(points)

思路2: 其实就是求最小生成树，首先想到的是prim 但是时间复杂度较高,超时

# prim算法 超出时间限制
class Solution:
    def minCostConnectPoints(self, points):
        # edge_list = []
        nodes = len(points)
        Matrix = [[0 for i in range(nodes)] for j in range(nodes)]
        for i in range(nodes):
            for j in range(nodes):
                dis = abs(points[i][0] - points[j][0]) + abs(points[i][1] - points[j][1])
                # edge_list.append([i, j, dis])
                Matrix[i][j] = dis
        # print('===edge_list:', edge_list)
        print('==Matrix:', Matrix)
        selected_node = [0]
        candidate_node = [i for i in range(1, nodes)]  # 候选节点
        print('==candidate_node:', candidate_node)
        # res = []
        res = 0
        while len(candidate_node):
            begin, end, minweight = 0, 0, float('inf')
            for i in selected_node:
                for j in candidate_node:
                    if Matrix[i][j] < minweight:
                        minweight = Matrix[i][j]
                        begin = i  # 存储开始节点
                        end = j  # 存储终止节点
            # res.append([begin, end, minweight])
            print('==end:', end)
            res += minweight
            selected_node.append(end)  # 找到权重最小的边　加入可选节点
            candidate_node.remove(end)  # 候选节点被找到　进行移除
        print('==res:', res)
        return res


points = [[0, 0], [2, 2], [3, 10], [5, 2], [7, 0]]
# points = [[-1000000, -1000000], [1000000, 1000000]]
sol = Solution()
sol.minCostConnectPoints(points)

思路3:并查集

class Solution:
    def find(self, x):
        if self.parents[x] == x:
            return x
        else:
            return self.find(self.parents[x])  # 一直找到帮主

    def union(self, i, j):  # 替换为帮主　站队
        self.parents[self.find(i)] = self.find(j)

    def minCostConnectPoints(self, points):
        costs_list = []
        n = len(points)
        for i in range(n):
            for j in range(i + 1, n):
                dis = abs(points[i][0] - points[j][0]) + abs(points[i][1] - points[j][1])
                costs_list.append([dis, i, j])
        # print('==costs_list:', costs_list)
        costs_list = sorted(costs_list, key=lambda x: x[0])
        print('==costs_list:', costs_list)

        self.parents = [i for i in range(n)]
        print('==init self.parents:', self.parents)
        res = 0
        for i in range(len(costs_list)):
            dis, x, y = costs_list[i]
            if self.find(x) != self.find(y):
                self.union(x, y)
                print('==x,y:', x, y)
                print('==self.parents:', self.parents)
                res += dis
        print('==res:', res)
        return res


points = [[0, 0], [2, 2], [3, 10], [5, 2], [7, 0]]
sol = Solution()
sol.minCostConnectPoints(points)

229.正则表达式匹配

思路:字符串的匹配问题，自然想到用ｄｐ，做成二维矩阵方便进行状态方程的转移

f[i][j]:s的前i个字符和p的前j个字符是否相等
p的第j个字符是字母：p[j]==s[i]时 ,f[i][j]=f[i-1][j-1]
p的第j个字符是字母:p[j]!=s[i]时 ,f[i][j]=False

p的第j个字符是*:s[i]==p[j-1],f[i][j]=f[i-1][j] or f[i][j-2]
p的第j个字符是*:s[i]!=p[j-1],f[i][j]=f[i][j-2]

#f[i][j]:s的前i个字符和p的前j个字符是否相等
#p的第j个字符是字母：p[j]==s[i]时 ,f[i][j]=f[i-1][j-1]
#p的第j个字符是字母:p[j]!=s[i]时 ,f[i][j]=False

#p的第j个字符是*:s[i]==p[j-1],f[i][j]=f[i-1][j] or f[i][j-2]
#p的第j个字符是*:s[i]!=p[j-1],f[i][j]=f[i][j-2]
class Solution:
    def match(self, i, j, s, p):
        if i == 0:
            return False
        if p[j - 1] == '.':
            return True
        return s[i - 1] == p[j - 1]

    def isMatch(self, s, p):
        m, n = len(s), len(p)

        f = [[False] * (n + 1) for _ in range(m + 1)]
        f[0][0] = True
        for i in range(m + 1):
            for j in range(1, n + 1):
                if p[j - 1] == '*':
                    # f[i][j] = f[i][j] or f[i][j - 2]
                    if self.match(i, j - 1, s, p):
                        f[i][j] = f[i - 1][j] or f[i][j-2]
                    else:
                        f[i][j] = f[i][j - 2]
                else:
                    if self.match(i, j, s, p):
                        f[i][j] = f[i - 1][j - 1]
                    else:
                        f[i][j] = False
        print('==f:', f)
        return f[m][n]


s = "aa"
p = "a"
sol = Solution()
sol.isMatch(s, p)

230.最长回文子串

思路:中心枚举，双指针，需要注意的是有上述两种情况，左右指针的索引有两种

class Solution:
    def help(self, left, right, s, n):
        while left >= 0 and right < n:
            if s[left] == s[right]:
                left -= 1
                right += 1
            else:
                break
        temp = s[left+1:right]
        self.res = max(self.res, temp, key=len)

    def longestPalindrome(self, s):
        self.res = s[0]
        n = len(s)
        for i in range(1, n):
            self.help(i-1, i+1, s, n)#针对"babad"
            self.help(i - 1, i, s, n)#针对"cbbd"
        print('==self.res:', self.res)
        return self.res

# s = "babad"
s = "cbbd"
sol = Solution()
sol.longestPalindrome(s)

231.寻找两个正序数组的中位数

思路:双指针，走完剩下的在进行合并

class Solution:
    def findMedianSortedArrays(self, nums1, nums2):
        res = []
        i, j = 0, 0
        m, n = len(nums1), len(nums2)
        while i < m and j < n:
            if nums1[i] < nums2[j]:
                res.append(nums1[i])
                i += 1
            else:
                res.append(nums2[j])
                j += 1
        print('==res:', res)
        print('==i:', i)
        print('==j:', j)
        if i < m:
            res.extend(nums1[i:])
        if j < n:
            res.extend(nums2[j:])
        print('==res:', res)
        if (m+n)%2==0:#偶数
            return (res[(m+n)//2]+res[(m+n)//2-1])/2
        else:#奇数
            return res[(m+n)//2]


# nums1 = [1, 1, 3]
# nums2 = [2]
nums1 = [1,2]
nums2 = [3,4]
sol = Solution()
res = sol.findMedianSortedArrays(nums1, nums2)
print(res)

232.连通网络的操作次数

思路:并查集 其实就是求当前的联通量个数减去一个联通量的值,对于n个节点,要给n-1条边才会满足都能连上,采用并查集去做聚类,否则就不满足了

class Solution:
    def find(self, x):
        if x==self.parent[x]:
            return x
        else:
            return self.find(self.parent[x])
    def union(self,x,y):#将x的老大换成y的老大
        self.parent[self.find(x)] = self.find(y)

    def makeConnected(self, n, connections):
        if len(connections) < n - 1:
            return -1
        self.parent = [i for i in range(n)]
        clusters = n
        for connection in connections:
           x, y = connection
           if self.find(x)!=self.find(y):
               clusters-=1
               self.union(x, y)
               print('==x,y', x, y)
               print('==self.parent:', self.parent)
        print(clusters)
        return clusters-1

n = 4
connections = [[0,1],
               [0,2],
               [1,2]]
sol = Solution()
sol.makeConnected(n, connections)

234.最长连续递增序列

思路:栈

class Solution:
    def findLengthOfLCIS(self, nums):
        stack = []
        res = 0
        for i in range(len(nums)):
            if stack and nums[i]<=stack[-1]:
                stack=[]
            stack.append(nums[i])
            res = max(len(stack), res)
            # print('==stack:', stack)
        # print(res)
        return res
nums = [1, 3, 5, 4, 7]
sol = Solution()
sol.findLengthOfLCIS(nums)

235.由斜杠划分区域

思路;把斜线换成3*3网格,就变成水域问题了

import numpy as np


class Solution:
    def dfs(self, i, j, h, w, matrix):
        if i < 0 or j < 0 or i >= h or j >= w or matrix[i][j] != 0:
            return
        matrix[i][j] = -1
        self.dfs(i - 1, j, h, w, matrix)
        self.dfs(i, j - 1, h, w, matrix)
        self.dfs(i + 1, j, h, w, matrix)
        self.dfs(i, j + 1, h, w, matrix)

    def regionsBySlashes(self, grid):
        n = len(grid)
        matrix = [[0 for _ in range(3 * n)] for _ in range(3 * n)]
        print(np.array(matrix))
        for i in range(n):
            for j in range(len(grid[i])):
                if grid[i][j] == '/':
                    matrix[i * 3][j * 3 + 2] = matrix[i * 3 + 1][j * 3 + 1] = matrix[i * 3 + 2][j * 3] = 1
                elif grid[i][j] == '\\':
                    matrix[i * 3 + 2][j * 3 + 2] = matrix[i * 3 + 1][j * 3 + 1] = matrix[i * 3][j * 3] = 1
        print(np.array(matrix))

        res = 0
        for i in range(3 * n):
            for j in range(3 * n):
                if matrix[i][j] == 0:
                    res+=1
                    self.dfs(i, j, 3 * n, 3 * n, matrix)
        print('==res:', res)
        return res

# grid = [" /", "/ "]
grid = ["/\\", "/\\"]
sol = Solution()
sol.regionsBySlashes(grid)

要注意的是如果格子用2*2,会出现这种0不会挨着的,会出错

236.等价多米诺骨牌对的数量

思路1;

两层循环 超时

class Solution:
    def numEquivDominoPairs(self, dominoes):
        #两层循环 超时
        n = len(dominoes)
        left, right = 0, 0
        res=0
        while left<n:
            right=left+1
            while right<n:
                if dominoes[right]==dominoes[left][::-1] or dominoes[right]==dominoes[left]:
                    res+=1
                right+=1
            left+=1
        print('==res:', res)
        return res

思路2:做成字典,记录相同的个数,两两之间相互组合为n*(n-1)/2

class Solution:
    def numEquivDominoPairs(self, dominoes):
        # 字典法
        ans = 0
        dict_ = {}
        for d1, d2 in dominoes:
            # 排序后加入字典
            index = tuple(sorted((d1, d2)))
            if index in dict_:
                dict_[index] += 1
            else:
                dict_[index] = 1
        print('==dict_:', dict_)
        # 计算答案
        for i in dict_:#n*n(-1)/2
            ans += dict_[i] * (dict_[i] - 1) // 2
        return ans

思路3:桶计数,两两之间相互组合为n*(n-1)/2

class Solution:
    def numEquivDominoPairs(self, dominoes):      
        #桶装法
        nums = [0]*100
        res = 0
        for dominoe in dominoes:
            x,y = dominoe
            if x>y:
                nums[x*10+y] +=1
            else:
                nums[y * 10 + x] += 1
        for num in nums:
            if num>=2:
                res += num*(num-1)//2
        print(res)
        return res

239.寻找数组的中心索引

class Solution:
    def pivotIndex(self, nums):
        n = len(nums)
        for i in range(n):
            left_sum = sum(nums[:i])
            right_sum = sum(nums[i+1:])
            if left_sum == right_sum:
                return i
        return -1

nums = [1, 7, 3, 6, 5, 6]
sol = Solution()
res = sol.pivotIndex(nums)
print(res)

240.最小体力消耗路径

思路1:咋一眼看过去,以为直接用dp就行,但是真正在写的过程中,发现消耗的最小体力没有状态转移方程,所以可不可以看成先初始化给一个体力值,如果能达到右下脚,将体力值减少,而这个减少的过程采用二分法这样减少更快

代码注释的部分使用list会超时,换成集合就好了,不会超时

# 思路:二分法,先给定一个初始体力值,能够走到的话,就减少体力,否则增加体力
class Solution:
    def minimumEffortPath(self, heights):
        h, w = len(heights), len(heights[0])
        left_value, right_value, res = 0, 10 ** 6 - 1, 0
        # 二分法
        while left_value <= right_value:  # 终止条件 找到合适的体力值
            mid = left_value + (right_value - left_value) // 2
            # start_point = [[0, 0]]
            # seen = [[0, 0]]
            start_point = [(0, 0)]
            seen = {(0, 0)}
            while start_point:
                x, y = start_point.pop(0)
                for x_, y_ in [(x - 1, y), (x + 1, y), (x, y - 1), (x, y + 1)]:
                    # 不超过范围 也没见过 差值也小与给定的体力值
                    # if 0 <= x_ < h and 0 <= y_ < w and [x_, y_] not in seen and abs(
                    #         heights[x][y] - heights[x_][y_]) <= mid:
                    #     start_point.append([x_, y_])
                    #     seen.append([x_, y_])
                    if 0 <= x_ < h and 0 <= y_ < w and (x_, y_) not in seen and abs(
                            heights[x][y] - heights[x_][y_]) <= mid:
                        start_point.append((x_, y_))
                        seen.add((x_, y_))
            print('==seen:', seen)
            if (h - 1, w - 1) in seen:  # 足够到底右下角,说明体力足够,那么继续减少体力
                res = mid
                right_value = mid - 1
            else:  # 到不了右下角,说明体力不够,需要增加体力
                left_value = mid + 1
        return res


heights = [[1, 2, 2],
           [3, 8, 2],
           [5, 3, 5]]
sol = Solution()
res = sol.minimumEffortPath(heights)
print('==res:', res)

思路2:并查集
将节点与节点之间的边与权重构建出来
按权重从小到大 不断合并 只要出现从起始点到终止点能够到达 说明找到了 此时的值就是最大消耗体力值

#思路:并查集
#将节点与节点之间的边与权重构建出来
#按权重从小到大 不断合并 只要出现从起始点到终止点能够到达 说明找到了 此时的值就是最大消耗体力值
class Solution:
    def find(self, x):
        if x==self.parent[x]:
            return x
        else:
            return self.find(self.parent[x])
    def union(self, x, y):
        #将y的老大换成x的老大
        x = self.find(x)
        y = self.find(y)
        if x == y:
            return False
        if self.size[x]<self.size[y]:
            x, y = y, x
        self.parent[y] = x
        self.size[x] += self.size[y]#x的老大进行计数 看哪边人多 那边人多 就归属
    def connect(self,x, y):
        x,y = self.find(x),self.find(y)
        return x==y

    def minimumEffortPath(self, heights):
        h, w = len(heights), len(heights[0])
        self.parent = [i for i in range(h*w)]
        self.size = [1 for _ in range(h*w)]

        edges = []
        #构建 节点和节点之间的边和权重
        for i in range(h):
            for j in range(w):
                index = i*w+j
                if i>0:
                    edges.append((index-w, index, abs(heights[i][j]-heights[i-1][j])))#减去一行的宽度 回到上一行 也就是竖线的边
                if j>0:
                    edges.append((index-1,index,abs(heights[i][j]-heights[i][j-1])))#减去左边一格  回到左边 也就是横线的边
        print('==edges:', edges)
        print(len(edges))
        edges = sorted(edges, key=lambda x: x[2])#按权重重小到大排序
        print('==edges:', edges)
        res = 0
        for x, y, value in edges:
            self.union(x, y)#把y的老大换成x的老大
            print('==self.parent:', self.parent)
            print('==self.size:', self.size)
            if self.connect(0, h*w-1):
                res = value
                break
        print('==res:', res)
        return res

heights = [[1, 2, 2],
           [3, 8, 2],
           [5, 3, 5]]
# heights  = [[4,3,4,10,5,5,9,2],
#             [10,8,2,10,9,7,5,6],
#             [5,8,10,10,10,7,4,2],
#             [5,1,3,1,1,3,1,9],
#             [6,4,10,6,10,9,4,6]]
sol = Solution()
res = sol.minimumEffortPath(heights)
print('==res:', res)

241.不用加减乘除做加法

思路:分为进位和 当前位的异或

class Solution:
    def add(self, a: int, b: int) -> int:
        add_flag = (a&b)<<1;#进位在左移
        sum_ = a^b;#异或 取值 相同为0 不同为1
        return sum_+add_flag

242.水位上升的泳池中游泳

堆队列 也称为优先队列 就是根节点的值最小

思路:采用bfs + 堆队列  保证能到达左下角的时候 需要最小的时间

# 堆队列 也称为优先队列 就是根节点的值最小
# 思路:采用dfs + 堆队列
import heapq


class Solution:
    def swimInWater(self, grid):
        h, w = len(grid), len(grid[0])
        res = 0
        heap = [(grid[0][0], 0, 0)]
        visited = {(0, 0)}

        while heap:
            print('===heap:', heap)
            height, x, y = heapq.heappop(heap)
            res = max(res, height)
            if x == w - 1 and y == h - 1:
                return res
            for dx, dy in ((0, 1), (0, -1), (1, 0), (-1, 0)):
                new_x, new_y = x + dx, y + dy
                if 0 <= new_x < w and 0 <= new_y < h and (new_x, new_y) not in visited:
                    visited.add((new_x, new_y))
                    heapq.heappush(heap, (grid[new_x][new_y], new_x, new_y))

        return -1

grid = [[0, 2],
        [1, 3]]
# grid = [[0, 1, 2, 3, 4],
#         [24, 23, 22, 21, 5],
#         [12, 13, 14, 15, 16],
#         [11, 17, 18, 19, 20],
#         [10, 9, 8, 7, 6]]
sol = Solution()
sol.swimInWater(grid)

243.从相邻元素对还原数组

思路:构建字典，进行bfs遍历 依次对出现过一次的字符 添加进列表即可

#bfs
class Solution():
    def restoreArray(self, adjacentPairs):
        """
        :type adjacentPairs: List[List[int]]
        :rtype: List[int]
        """
        import collections
        memory = collections.defaultdict(list)
        for x,y in adjacentPairs:
            memory[x].append(y)
            memory[y].append(x)
        print('==memory:', memory)
        res = []
        visited = set()
        queue = []
        for key, value in memory.items():
            if len(value) == 1:
                queue.append((key, value[0]))
        while queue:
            print('==queue:', queue)
            start, end = queue.pop()
            res.append(start)
            print('==res:', res)
            visited.add(start)
            for num in memory[end]:
                if num in visited:
                    continue
                queue.append((end, num))
        return res

adjacentPairs = [[2,1],
                 [3,4],
                 [3,2]]
# # adjacentPairs = [[4,-2],
# #                  [1,4],
# #                  [-3,1]]
# # adjacentPairs = [[100000,-100000]]
# adjacentPairs =[[4,-10],
#                 [-1,3],
#                 [4,-3],
#                 [-3,3]]
sol = Solution()
ress= sol.restoreArray(adjacentPairs)
print(ress)

246.公平的糖果棒交换

#思路1:两层for循环 超时

#思路1:两层for循环 超时
class Solution:
    def fairCandySwap(self, A, B):
        sum_A = sum(A)
        sum_B = sum(B)
        target = (sum_A+sum_B)//2
        for i in range(len(A)):
            for j in range(len(B)):
                if sum(A[:i])+sum(A[i+1:])+B[j] == target:
                    return [A[i], B[j]]
        return []

# A = [1, 2]
# B = [2, 3]
A = [1 ,2, 5]
B = [2, 4]
sol = Solution()
res = sol.fairCandySwap(A, B)
print(res)

 思路2:双指针 利用 sumA-x+y = sumB+x-y -->(sumA-sumB)/2 = x-y  转换为寻找两个元素

#思路1:#sumA-x+y = sumB+x-y -->(sumA-sumB)/2 = x-y
# 故用双指针 转换为二分查找 去查找两数之差
class Solution:
    def fairCandySwap(self, A, B):
        i = 0
        j = 0
        A = sorted(A)
        B = sorted(B)
        sum_A  = sum(A)
        sum_B = sum(B)
        target = (sum_A-sum_B)/2
        print('==target:', target)
        while i<len(A) and j<len(B):
            if (A[i] - B[j]) == target:
                return [A[i], B[j]]
            elif (A[i] - B[j])>target:
                j+=1
            else:
                i+=1
        return []

# A = [1, 2]
# B = [2, 3]
A = [1,2,5]
B = [2,4]
sol = Solution()
res = sol.fairCandySwap(A, B)
print(res)

思路3:一层for循环 利用 sumA-x+y = sumB+x-y -->(sumA-sumB)/2 = x-y  转换为寻找两个元素

#思路2:#sumA-x+y = sumB+x-y -->(sumA-sumB)/2 = x-y
# 一层for循环走遍
class Solution:
    def fairCandySwap(self, A, B):
        sum_A = sum(A)
        sum_B = sum(B)
        target = (sum_A-sum_B)//2
        for num in B:
            if num+target in A:
                return [num+target, num]
        return []

A = [1, 2]
B = [2, 3]
# A = [1 ,2, 5]
# B = [2, 4]
sol = Solution()
res = sol.fairCandySwap(A, B)
print(res)

247.替换后的最长重复字符

思路:如果k==0其实就是求最长字符串 ,k大于0,那么采用滑动窗口的方式 对左右指针求最少字符(也可能是不同字符)个数

如果最少字符个数大于k说明 左指针该右移动了,否则右指针一直在右移

# 思路:如果k==0其实就是求最长字符串 ,k大于0,那么采用滑动窗口的方式 对左右指针求最少字符个数
# 如果最少字符个数大于k说明 左指针改右移动了,否则右指针一直在右移
class Solution:
    def characterReplacement(self, s, k):
        left, right = 0, 0
        nums = [0] * 26
        # max_num = 0
        res = 0
        while right < len(s):
            nums[ord(s[right]) - ord('A')] += 1
            max_num = max(nums)
            #减去max_num 就是 求左右指针内最少字符(也可能是不同字符)个数
            while right - left + 1 - max_num > k:
                nums[ord(s[left]) - ord('A')] -= 1
                left += 1
            right += 1
            res = max(right - left, res)
        print('==res:', res)
        return res


# s = "ABAB"
# k = 2
s = "AABABBA"
k = 1
sol = Solution()
res = sol.characterReplacement(s, k)
print('==res:', res)

250.子数组最大平均数 I

思路1:采用list和 ,超时

class Solution:
    def findMaxAverage(self, nums, k):
        res = float('-inf')
        for i in range(len(nums)-k+1):
            res = max(sum(nums[i:i+k]), res)
        print(res)
        return res/k

nums = [1, 12, -5, -6, 50, 3]
k = 4
sol = Solution()
sol.findMaxAverage(nums, k)

思路2:采用前缀和,空间换时间

class Solution:
    def findMaxAverage(self, nums, k):
        length =len(nums)
        pre_sum = [0]*length
        pre_sum[0] = nums[0]
        for i in range(1, length):
            pre_sum[i] = pre_sum[i-1]+nums[i]
        print('===pre_sum:', pre_sum)

        res = float('-inf')
        for i in range(k-1, length):
            if i > k-1:
                res = max(res, pre_sum[i] - pre_sum[i-k])
            else:
                res = pre_sum[i]
        print('==res:', res)
        return res/k


nums = [1, 12, -5, -6, 50, 3]
k = 4
sol = Solution()
res = sol.findMaxAverage(nums, k)
print('==res:', res)

思路3:双指针

class Solution:
    def findMaxAverage(self, nums, k):
        length = len(nums)
        Sum = 0
        res = float('-inf')
        left,right=0,0
        while right<length:
            Sum += nums[right]
            if right>=k-1:
                res =max(res, Sum)
                print('==res:', res)
                Sum-=nums[left]
                left += 1
            right+=1
        return res/k

nums = [1, 12, -5, -6, 50, 3]
k = 4
sol = Solution()
res = sol.findMaxAverage(nums, k)
print('==res:', res)

251.尽可能使字符串相等

思路;转换成距离列表后进行双指针滑动窗口即可

#双指针
class Solution:
    def equalSubstring(self, s, t, maxCost):
        length = len(s)
        dist_cost = [0] * length
        for i in range(length):
            dist_cost[i] = abs(ord(s[i]) - ord(t[i]))
        print('==dist_cost:', dist_cost)
        res = 0
        left,right =0,0
        Sum = 0
        while right<length:
            Sum+=dist_cost[right]
            if Sum>maxCost:
                Sum -= dist_cost[left]
                left+=1
            right+=1
            res = max(res, right - left)
        print(res)
        return res
# s = "abcd"
# t = "bcdf"
# cost = 3
# s = "abcd"
# t = "acde"
# cost = 0
s = "krrgw"
t = "zjxss"
cost = 19
# s = "abcd"
# t = "cdef"
# cost = 3
sol = Solution()
sol.equalSubstring(s, t, cost)

261.可获得的最大点数

思路:转换为求连续的和最小, 那么自然用滑动窗口解决即可以

class Solution:
    def maxScore(self, cardPoints, k):
        length = len(cardPoints)
        leaving_k = length - k
        print('==leaving_k:', leaving_k)
        if leaving_k == 0:
            return sum(cardPoints)

        left, right = 0, 0
        min_res = float('inf')
        temp_res = 0
        while right < length:
            temp_res += cardPoints[right]
            if right >= leaving_k - 1:
                min_res = min(min_res, temp_res)
                print('==temp_res:', temp_res)
                left += 1
                temp_res -= cardPoints[left - 1]
            right += 1
        print('==min_res:', min_res)
        return sum(cardPoints) - min_res

cardPoints = [1, 2, 3, 4, 5, 6, 1]
k = 3
# cardPoints = [9, 7, 7, 9, 7, 7, 9]
# k = 7
sol = Solution()
sol.maxScore(cardPoints, k)

262.非递减数列

思路:分别从左右两边判断是否递增

class Solution:
    def checkPossibility(self, nums: List[int]) -> bool:
        length = len(nums)
        # res = 0
        left,right = 0,length-1
        while left<length-1 and nums[left]<=nums[left+1]:
            left+=1
        if left==length-1:
            return True
        while right>=0 and nums[right-1]<=nums[right]:
            right-=1
        if right - left>1:
            return False
        if left==0 or right==length-1:
            return True
        
        if nums[right+1]>=nums[left] or nums[left-1]<=nums[right]:
            return True

        return False

264.最长湍流子数组

思路:双指针

满足山峰:arr[right]>arr[right-1] and arr[right]>arr[right+1]  right+=1

满足山谷:arr[right]<arr[right-1] and arr[right]<arr[right+1]  right+=1

其他时候 left移动到right位置

class Solution:
    def maxTurbulenceSize(self, arr):
        left, right = 0, 0
        length = len(arr)
        res = 1
        while right<length-1:
            if left==right:
                if left+1<length and arr[left]==arr[left+1]:
                    left+=1
                right+=1
            else:#山峰
                if right+1<length and arr[right]>arr[right-1] and arr[right]>arr[right+1]:
                    right+=1
                # 山谷
                elif right+1<length and arr[right]<arr[right-1] and arr[right]<arr[right+1]:
                    right+=1
                else:
                    left=right
            print('==right:', right)
            res = max(res, right-left+1)
        print('==res:', res)
        return res


# arr  = [9,4,2,10,7,8,8,1,9]
# arr = [100]
arr = [2,1]
sol = Solution()
sol.maxTurbulenceSize(arr)

267.数据流中的第 K 大元素

思路1:每个add进去 就sort取第k大，时间复杂度偏大k*log(k),对于这种取topk问题，用最小堆更合适

#k*O(logk)  超时
class KthLargest:

    def __init__(self, k, nums):
        self.k = k
        self.nums = nums

    def add(self, val):
        self.nums.append(val)
        self.nums = sorted(self.nums)[::-1]
        return self.nums[self.k - 1]
k = 3
nums = [4, 5, 8, 2]
sol = KthLargest(k, nums)
res = sol.add(3)
print('=res:', res)
res = sol.add(5)
print('=res:', res)
res = sol.add(10)
print('=res:', res)
res = sol.add(9)
print('=res:', res)
res = sol.add(4)
print('=res:', res)

思路2:最小堆，保证最小堆中只有k个元素，那么堆顶自然就是第k大元素

时间复杂度为log(k),因为push和pop都是log(k).

python代码

# 最小堆 topk都用最小堆
import heapq
class KthLargest(object):

    def __init__(self, k, nums):
        """
        :type k: int
        :type nums: List[int]
        """
        self.k = k
        self.que = nums
        heapq.heapify(self.que)

    def add(self, val):
        """
        :type val: int
        :rtype: int
        """
        heapq.heappush(self.que, val)
        print('=====self.que====:', self.que)
        while len(self.que)>self.k:#保持最小堆中只有k个元素 则堆顶就是第k大元素
            heapq.heappop(self.que)
        print('clean self.que:', self.que)
        return self.que[0]

k = 3
nums = [4, 5, 8, 2]
sol = KthLargest(k, nums)
res = sol.add(3)
print('=res:', res)
res = sol.add(5)
print('=res:', res)

c++代码:

#include <map>
#include <vector>
#include <iostream>
#include <queue>
using namespace std;

class KthLargest {
public:
    priority_queue<int, vector<int>, greater<int> > que;//最小堆
    int k;
    KthLargest(int k, vector<int>& nums) {
        this->k = k;
        for(int k=0;k<nums.size();k++)
        {
            que.push(nums[k]);
            if (que.size()>this->k)
            {
                que.pop();
            }
        }
    }
    
    int add(int val) {
        que.push(val);
        if(que.size()>this->k)
        {
            que.pop();
        }
        return que.top();
    }
};

int main()
{   
    int k=3;
    vector<int> nums;
    nums={4,5,8,2};
    KthLargest *p = new KthLargest(k, nums);
    int res = p->add(3);
    cout<<"res:"<<res<<endl;
    res = p->add(5);
    cout<<"res:"<<res<<endl;
    res = p->add(10);
    cout<<"res:"<<res<<endl;
    res = p->add(9);
    cout<<"res:"<<res<<endl;
    res = p->add(4);
    cout<<"res:"<<res<<endl;

    delete p;
    p=NULL;
    return 0;

}

269.杨辉三角 II

思路:一层一层遍历出值即可

python代码:

class Solution:
    def getRow(self, rowIndex: int) -> List[int]:
        temp = [0]*(rowIndex+1)
        temp[0] = 1
        for i in range(1, rowIndex+1):
            for j in range(i, 0, -1):
                temp[j] += temp[j-1]
            # print('==temp:', temp)
        return temp

c++代码:

#include <map>
#include <vector>
#include <iostream>
#include <queue>
#include <algorithm>
using namespace std;
class Solution {
public:
    vector<int> getRow(int rowIndex) {
        vector<int> temp(rowIndex+1, 0);
        temp[0] = 1;
        for (int i=1;i<rowIndex+1;i++)
        {
            for(int j=i;j>0;j--)
            {
                temp[j]+=temp[j-1];
            } 
        }
        return temp;
    }
};

int main()
{   
    Solution *p = new Solution();
    int row_index = 3;
    vector <int> res;
    res = p->getRow(row_index);
    for(int k=0;k<res.size();k++)
    {
        cout<<"res[k]"<<res[k]<<endl;
    }
    p=NULL;
    delete p;
    return 0;
}

271.找到所有数组中消失的数字

思路1:hash  空间复杂度 o(n) 时间复杂度o(n)

# 空间复杂度 o(n) 时间复杂度o(n)
class Solution:
    def findDisappearedNumbers(self, nums):
        dict_={}
        for i in range(len(nums)):
            dict_[nums[i]] = dict_.get(nums[i],0)+1

        res = []
        for i in range(len(nums)):
            if i+1 not in dict_:
                res.append(i+1)
        print(res)
        return res

思路2:求出索引在对应位置处 添加长度 如果没有的数字,则数字就小于等于长度

空间复杂度O(1) 时间复杂度O(n)

#空间复杂度O(1) 时间复杂度O(n)
class Solution:
    def findDisappearedNumbers(self, nums):
        length = len(nums)
        for i in range(len(nums)):
            index= (nums[i]-1)%length
            nums[index] +=length
        print('==nums:', nums)
        res = [i+1 for i in range(length) if nums[i] <= length]
        print(res)
        return res

# nums = [4,3,2,7,8,2,3,1]
nums = [1,2,2,3,3,4,7,8]
# nums = [1,2,3,3,5,6,7]
sol = Solution()
sol.findDisappearedNumbers(nums)

c++代码:

#include <map>
#include <vector>
#include <iostream>
#include <queue>
#include <string>
#include <algorithm>
using namespace std;
class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) {
        vector<int> res;
        for (int i=0;i<nums.size();i++)
        {
            int index = (nums[i]-1)%nums.size();
            nums[index] += nums.size();
        }
        for (int i=0;i<nums.size();i++)
        {
            if(nums[i]<=nums.size())
            {
                res.push_back(i+1);
            }
        }
        return res;
    }
};

int main()
{
    Solution *p = new Solution();
    vector <int >nums;
    nums = {4,3,2,7,8,2,3,1};
    vector <int> res = p->findDisappearedNumbers(nums);
    for(int i=0;i<res.size();i++)
    {
        cout<<"res:"<<res[i]<<endl;    
    }
    
    delete p;
    p = NULL;
    return 0;
}

 272.情侣牵手

思路:其实就是将环拆开，0,1都看成第0对，2,3看成第1对

可看出要交换的座位就是环的边数减去1,对于这种去环问题，自然想到并查集

python代码:

class Solution:
    def find(self,x):
        if self.parent[x]==x:
            return x
        return self.find(self.parent[x])
    def union(self,i,j):#将i的老大变成j的老大
        self.parent[self.find(i)] = self.find(j)
    def get_count(self,n):
        for i in range(n):
            self.count[self.find(i)]+=1
    def minSwapsCouples(self, row):
        n = len(row)//2
        self.parent = [i for i in range(n)]
        self.count = [0 for i in range(n)]
        print('===init self.parent', self.parent)
        for i in range(0, len(row), 2):
            self.union(row[i]//2, row[i+1]//2)
            print('==self.parent:', self.parent)
        self.get_count(n)
        print('===self.count:', self.count)
        res = 0
        for i in range(n):
            res += max(self.count[i]-1, 0)
        print(res)
        return res

# row = [0,2,2]
# row = [0, 2, 1, 3]
# row = [2,0,5,4,3,1]
row = [1,4,0,5,8,7,6,3,2,9]
# row = [0, 1, 2, 3]
sol = Solution()
sol.minSwapsCouples(row)

c++代码:

#include <map>
#include <vector>
#include <iostream>
#include <queue>
#include <string>
#include <algorithm>
using namespace std;

class Solution {
public:
    vector<int> parent;
    vector<int> count;
    int find(int x)
    {
        if(x==parent[x])
        {
            return x;
        }
        return find(parent[x]);
    }
    //把i的老大换成j的老大
    void merge(int i, int j)
    {
        parent[find(i)]=find(j);
    }
    void get_count(int n)
    {
        for (int i=0;i<n;i++)
        {   
            count[find(i)]+=1;
        }
    }
    int minSwapsCouples(vector<int>& row) {
        int n = row.size()/2;
        cout<<n<<endl;
        int res=0;
        for(int i=0;i<n;i++)
        {
            parent.push_back(i);
            count.push_back(0);
        }
        for(int i=0;i<row.size();i+=2)
        {
            merge(row[i]/2,row[i+1]/2);
        }
        get_count(n);
        // // //debug 
        // for (int i=0;i<parent.size();i++)
        // {
        //     cout<<"===parent[i]"<<parent[i]<<endl;
        // }
        //debug 
        // for (int i=0;i<count.size();i++)
        // {
        //     cout<<"===count[i]"<<count[i]<<endl;
        // }
        for (int i=0;i<count.size();i++)
        {
            res+=max(count[i]-1,0);
        }

        return res;
    }
};
int main()
{
    Solution *p = new Solution();
    vector<int> row;
    row = {0, 2, 1, 3};
    int res = p->minSwapsCouples(row);
    cout<<"==res:"<<res<<endl;
    delete p;
    p=NULL;
    return 0;
}

273.最大连续1的个数

思路1:直接数1个数

class Solution:
    def findMaxConsecutiveOnes(self, nums):
        count_one = 0
        res = 0
        for i in range(len(nums)):
            if nums[i]==1:
                count_one+=1
            else:
                count_one=0
            res = max(res, count_one)
        # print(res)
        return res

思路2：dp

class Solution:
    def findMaxConsecutiveOnes(self, nums):
        res = [-1]
        for i in range(len(nums)):
            if nums[i]==0:
                res.append(i)
        res.append(len(nums))
        print(res)
        if len(res)==1:
            return res[-1]
        max_length = 0
        for i in range(1, len(res)):
            max_length = max(res[i]-res[i-1]-1, max_length)
        print(max_length)
        return max_length

思路3：双指针滑动窗口

class Solution:
    def findMaxConsecutiveOnes(self, nums):
        left,right=0,0
        res = 0
        while right<len(nums):
            if nums[right]==1:
                right+=1
            else:
                right+=1
                left=right
            print('==left,right:', left, right)
            res = max(res, right - left)
            print('==res:', res)
        return res

c++双指针:

#include <vector>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

class Solution {
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
        int left=0;
        int right = 0;
        int res = 0;
        while (right<nums.size())
        {
            if(nums[right]==1)
            {
                right++;
            }
            else
            {                   
                right++;
                left=right;                
            }
            res = max(right-left, res);
        }
        return res;        
    }
};

int main()
{
    Solution *p = new Solution();
    vector<int> nums;
    nums = {1,1,0,1,1,1};
    int res = p->findMaxConsecutiveOnes(nums);
    cout<<"==res:"<<res<<endl;
    delete p;
    p=NULL;
    return 0;
}

274.重塑矩阵

思路:对于h*w的元素个数，索引为h_index = i//rows,w_index = i%rows

python代码

class Solution:
    def matrixReshape(self, nums: List[List[int]], r: int, c: int) -> List[List[int]]:
        h = len(nums)
        w = len(nums[0])
        if h*w != r*c:
            return nums

        res = [[0 for _ in range(c)] for _ in range(r)]
        # print(res)
        for i in range(h*w):
            res[i//c][i%c] = nums[i//w][i%w]
        # print(res)
        return res

c++代码:

class Solution {
public:
    vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) {
        int h = nums.size();
        int w = nums[0].size();
        // cout<<"==h:"<<h<<endl;
        vector<vector<int>> res(r ,vector<int>(c,0));  //初始化c*r元素为0的矩阵 
        
        // cout<<"==res.size():"<<res.size()<<endl;
        // cout<<"==res[0].size():"<<res[0].size()<<endl;
        if(h*w!=r*c)
        {
            return nums;
        }
        for (int i=0;i<h*w;i++)
        {   
            // cout<<"i/w:"<<i/w<<endl;
            // cout<<"i%w:"<<i%w<<endl;
            res[i/c][i%c] = nums[i/w][i%w];
        }
        return res;
    }
};

275.最大连续1的个数 III

思路:其实就是滑动窗口判断有大于K个0的则左指针右移动，之所以用大于K来判断是因为0后续可能会跟着很多1,所以大于K的话，会把这些包含进去,和https://leetcode-cn.com/problems/longest-substring-without-repeating-characters/ 思想一样

python:

class Solution:
    def longestOnes(self, A: List[int], K: int) -> int:
        left,right =0,0
        n = len(A)
        zero_count =0
        res = 0
        while right<n:
            if A[right]==0:
                zero_count+=1
            #left向右收缩
            while zero_count>K:#大于K个0的时候就说明找到符合条件的了
                if A[left]==0:
                    zero_count-=1
                left+=1
            res = max(res, right - left + 1)
            # print('=left:', left)
            # print('==right:', right)
            # print('==res:', res)
            right+=1
        # print(res)
        return res

c++

class Solution {
public:
    int longestOnes(vector<int>& A, int K) {
        int left=0;
        int right =0;
        int res=0;
        int zero_count =0;
        int length = A.size();
        while(right<length)
        {   
            if (A[right]==0)
            {
                zero_count++;
            }
            while (zero_count>K)
            {   
                if (A[left]==0)
                {
                    zero_count--;
                }
                left++;
            }
            res = max(res, right-left+1);
            
            right++;
        }
        return res;

    }
};

276.数组的度

思路:三个hash,一个计度，一个记录左索引，一个记录右索引

python:

class Solution:
    def findShortestSubArray(self, nums: List[int]) -> int:
        dict_ = {}
        left_index_dict = {}
        right_index_dict = {}
        for i in range(len(nums)):
            dict_[nums[i]] = dict_.get(nums[i], 0)+1
            if nums[i] not in left_index_dict:
                left_index_dict[nums[i]] = i
            right_index_dict[nums[i]] = i
        # print('==dict_:', dict_)
        # print('=left_index_dict:', left_index_dict)
        # print('==right_index_dict:', right_index_dict)
        dict_ = dict(sorted(dict_.items(), key=lambda x:x[1],reverse=True))
        # print(dict_)
        max_degree = 0
        res = float('inf')
        for key,value in dict_.items():
            max_degree = value
            break
        # print(max_degree)
        for key, value in dict_.items():
            if value==max_degree:
                res = min(res, right_index_dict[key] - left_index_dict[key]+1)
        # print('===res:',res)
        return res

c++:

class Solution {
public:
    int findShortestSubArray(vector<int>& nums) {
        map<int, int>dict_;
        map<int, int>left_index_dict;
        map<int, int>right_index_dict;
        for(int i=0;i<nums.size();i++){
            dict_[nums[i]]++;
            if(left_index_dict.count(nums[i])==0){
                left_index_dict[nums[i]] = i;
            }
            right_index_dict[nums[i]] = i;
        }
        int max_degree=0;
        map<int,int>::iterator iter=dict_.begin();
        for (;iter!=dict_.end();iter++)
        {
            max_degree = max(max_degree, iter->second);
        }
        int res=INT_MAX;
        map<int,int>::iterator iter_2=dict_.begin();
        for (;iter_2!=dict_.end();iter_2++)
        {
            if (max_degree == iter_2->second){
                res = min(res, right_index_dict[iter_2->first] - left_index_dict[iter_2->first]+1);
            }
        }
        return res;

    }
};