[数据结构与算法]第 287 场周赛

第 287 场周赛

6055. 转化时间需要的最少操作数

class Solution:
    def convertTime(self, current: str, correct: str) -> int:
        d = int(correct[:2]) * 60 + int(correct[3:]) - (int(current[:2]) * 60 + int(current[3:]))
        res = 0
        for i in [60, 15, 5, 1]:
            res += d // i
            d %= i
        return res

5235. 找出输掉零场或一场比赛的玩家

class Solution:
    def findWinners(self, matches: List[List[int]]) -> List[List[int]]:
        win, los = defaultdict(int), defaultdict(int)
        for a, b in matches:
            win[a] += 1
            los[b] += 1
        w = set(win) - set(los)
        u = [x for x, v in los.items() if v == 1]
        return [sorted(w), sorted(u)]

2226. 每个小孩最多能分到多少糖果

class Solution {
    public int maximumCandies(int[] candies, long k) {
        long sum = 0L; 
        for(int i : candies) sum += i;
        if(sum < k) return 0;
        if(sum == k) return 1;
        long i = 1, j = sum / k;
        while (i <= j){  
            long mid = (i + j)/2;
            long cnt = 0L;
            for(int num : candies){
                cnt += num / mid;
            }
            if(cnt >= k) i = mid + 1;
            else j = mid - 1;
        }
        return (int)i - 1;
    }
}

2227. 加密解密字符串