DW&LeetCode_day4(1018、16、20、21)
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table of Contents
DW&LeetCode_day4(1018、16、20、21)
1018. Binary prefix divisible by 5
16. The closest sum of three numbers
21. Merge two ordered linked lists
1018. Binary prefix divisible by 5
Given an array A consisting of a number of 0s and 1s. We define N_i: the i-th sub-array from A[0] to A[i] is interpreted as a binary number (from the most significant bit to the least significant bit).
Return a list of boolean values answer, only when N_i is divisible by 5, answer[i] is true, otherwise it is false.
Example 1:
Input: [0,1,1]
Output: [true,false,false]
Explanation: The
input numbers are 0, 01, 011; that is, 0, 1, 3 in decimal. Only the first number is divisible by 5, so answer[0] is true.Example 2:
Input: [1,1,1]
Output: [false,false,false]
Example 3:Input: [0,1,1,1,1,1]
Output: [true,false,false,false,true,false]
Example 4:Input: [1,1,1,0,1]
Output: [false,false,false,false,false]
prompt:
1 <= A.length <= 30000
A[i] 为 0 或 1Link: topic link
answer:
class Solution: def prefixesDivBy5(self, A: List[int]) -> List[bool]: a = [] b = 0 for i in A: b = (b * 2 + i)% 5 #将前面的数乘2，再加上新增数并对5取余 if b == 0: a+=True else: a+=False return a
16. The closest sum of three numbers
Given an array of n integers nums and a target value target. Find the three integers in nums so that their sum is closest to target. Returns the sum of these three numbers. Assume that there is only one answer for each set of inputs.
Example:
Input: nums = [-1,2,1,-4], target = 1
Output: 2
Explanation: The closest sum to target is 2 (-1 + 2 + 1 = 2).
prompt:
3 <= nums.length <= 10^3
-10^3 <= nums[i] <= 10^3
-10^4 <= target <= 10^4
Link: topic linkanswer:
# python 双指针加排序 class Solution: def threeSumClosest(self, a: List[int], target: int) -> int: a.sort() res = float('inf') for i in range(len(a)): if i == 0 or a[i]>a[i-1]: # 去重 left = i+1 right = len(a)-1 while left<right: b = a[i]+a[left]+a[right] if b == target: return target if abs(b-target)<abs(res-target): # 更新 res = b if b < target: left+=1 else: right-=1 return res
20. Valid parentheses
Given a string that only includes'(',')','{','}','[',']', judge whether the string is valid.
A valid string must meet:
The opening bracket must be closed with the same type of closing bracket.
The opening parenthesis must be closed in the correct order.
Note that an empty string can be considered a valid string.Example 1:
Input: "()"
Output: true
Example 2:Input: "()[]{}"
Output: true
Example 3:Input: "(]"
Output: false
Example 4:Input: "([)]"
Output: false
Example 5:Input: "{[]}"
Output: true
Link: topic linkanswer:
# python 语法糖 while '{}' in s or '()' in s or '[]' in s: s = s.replace('{}', '') s = s.replace('[]', '') s = s.replace('()', '') return s == ''
21. Merge two ordered linked lists
Combine two ascending linked lists into a new ascending linked list and return. The new linked list is composed by splicing all the nodes of the given two linked lists.
Example 1:
Input: l1 = [1,2,4], l2 = [1,3,4]
Output: [1,1,2,3,4,4]
Example 2:Input: l1 = [], l2 = []
Output: []
Example 3:Input: l1 = [], l2 = [0]
Output: [0]
Link: topic linkanswer:
# 一个详细的思路 --alex class Solution: def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode: #创建一个新的链表 res = ListNode(0) #一个指针指向新的链表的头部 c1 = res #开始比较L1和l2的值，只要有一个走完就停 while(l1 and l2): #如果l1的值小于等于l2的值，则加入新的链表同时移动L1 if l1.val<=l2.val: c1.next = ListNode(l1.val) l1 = l1.next c1 = c1.next #反之则添加l2的值，同时移动l2 else l1.val>l2.val: c1.next = ListNode(l2.val) l2= l2.next c1 =c1.next #检查L1和l2哪个没走完，则将后面的直接加入新的链表即可 if l1: c1.next = l1 if l2: c1.next = l2 #返回头部的下一个即可 return res.nextclass Solution: def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode: if not l1 : return l2 #终止条件 if not l2 : return l1 if l1.val <= l2.val : #递归条件 return ListNode(l1.val, self.mergeTwoLists(l1.next, l2)) else: return ListNode(l2.val, self.mergeTwoLists(l1, l2.next))
