【leetcode系列】【算法】第 185 场周赛

题目一：重新格式化字符串

题目链接： https://leetcode-cn.com/problems/reformat-the-string/

解题思路：

先遍历字符串，把字母和数字分开保存在两个容器中

从数目比较多的开始交替遍历拼接字符串

如果最后少的已经用完了，多的还有剩余，则表示无法重新格式化，返回空字符串

代码实现：

class Solution:
    def reformat(self, s: str) -> str:
        rec = collections.defaultdict(list)
        lst1 = 'qwertyuiopasdfghjklzxcvbnm'
        lst2 = '1234567890'
        for a in s:
            if a in lst1:
                rec['a'].append(a)
            else:
                rec['b'].append(a)
                
        if len(rec['a']) < len(rec['b']):
            rec['a'], rec['b'] = rec['b'][:], rec['a'][:]
        res = ''
        while 0 < len(rec):
            if 0 == len(rec['a']):
                if 0 != len(rec['b']):
                    return ''
                return res
            
            res += rec['a'][-1]
            rec['a'].pop()
            
            if 0 == len(rec['b']):
                if 0 != len(rec['a']):
                    return ''
                
                return res
            res += rec['b'][-1]
            rec['b'].pop()
            
        return ''

题目二：点菜展示表

题目链接： https://leetcode-cn.com/problems/display-table-of-food-orders-in-a-restaurant/

解题思路：

先遍历订单，合并同一桌的所有菜，同时保存所有点过的菜

然后将所有点过的菜名排序，并对排序后的桌号进行遍历拼接对应菜品的个数

代码实现：

class Solution:
    def displayTable(self, orders: List[List[str]]) -> List[List[str]]:
        rec = collections.defaultdict(list)
        food_lst = set()
        for content in orders:
            rec[int(content[1])].append(content[2])
            food_lst.add(content[2])
                    
                
        food_lst = list(food_lst)
        food_lst.sort()
        
        res = [['Table'] + food_lst]
        for table_id in sorted(rec.keys()):
            curr_content = [str(table_id)]
            for food in food_lst:
                curr_content.append(str(rec[table_id].count(food)))
                
            res.append(curr_content)
        return res

题目三：数青蛙

题目链接： https://leetcode-cn.com/problems/minimum-number-of-frogs-croaking/

解题思路：

添加一个变量保存喊完一次的青蛙个数

如果没有青蛙的声音结束，则来新的c时，新增青蛙个数

否则，更新已经喊完的青蛙个数 - 1，继续遍历

代码实现：

class Solution:
    def minNumberOfFrogs(self, croakOfFrogs: str) -> int:
        num = 0
        finish_num = 0
        last_word_tab = {'r' : 'c', 'o' : 'r', 'a' : 'o', 'k' : 'a'}
        res = {'c' : 0, 'r' : 0, 'o' : 0, 'a' : 0, 'k' : 0}
        for word in croakOfFrogs:
            if word == 'c':
                if finish_num <= 0:
                    num += 1
                else:
                    finish_num -= 1
                is_add_new = True
                res['c'] += 1
                continue
            elif word == 'k':
                if res['a'] > 0:
                    res['a'] -= 1
                    finish_num += 1
                    continue
                else:
                    return -1
                    
            res[last_word_tab[word]] -= 1
            res[word] += 1
            if res[last_word_tab[word]] < 0:
                return -1
            
        if sum(res.values()) > 0:
            return -1
        
        return num

题目四：

题目链接： https://leetcode-cn.com/problems/build-array-where-you-can-find-the-maximum-exactly-k-comparisons/

解题思路：

三维动态规划，没做出来，详细看大神的题解

代码实现：

class Solution:
    def numOfArrays(self, n: int, m: int, k: int) -> int:
        if k == 0:
            return 0
        dp = [[[0] * (m) for _ in range(k + 1)] for __ in range(n)]
        dp[0][1] = [1] * m
        
        for i in range(1, n):
            for j in range(1, k + 1):
                for p in range(m):
                    dp[i][j][p] = sum(dp[i - 1][j - 1][: p]) + dp[i - 1][j][p] * (p + 1)
                    dp[i][j][p] %= 1e9+7
                    
        return int(sum(dp[-1][-1])%(1e9+7))

作者：coldme-2
链接：https://leetcode-cn.com/problems/build-array-where-you-can-find-the-maximum-exactly-k-comparisons/solution/jian-dan-san-wei-dong-tai-gui-hua-by-coldme-2/
来源：力扣（LeetCode）
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