Leetcode 539. 最小时间差

给定一个 24 小时制（小时:分钟 “HH:MM”）的时间列表，找出列表中任意两个时间的最小时间差并以分钟数表示。

示例 1：

输入：timePoints = [“23:59”,“00:00”]
输出：1
示例 2：

输入：timePoints = [“00:00”,“23:59”,“00:00”]
输出：0

提示：

2 <= timePoints.length <= 2 * 104
timePoints[i] 格式为 “HH:MM”

class Solution {
public:
    int findMinDifference(vector<string>& timePoints) {
        vector<int> time;
        for(auto s : timePoints) {
            int h, m;
            if(s[0] == '0') h = s[1] - '0';
            else h = (s[0] - '0') * 10 + (s[1] - '0');
            if(s[3] == '0') m = s[4] - '0';
            else m = (s[3] - '0') * 10 + (s[4] - '0');

            time.push_back(h * 60 + m);
        }
        sort(time.begin(), time.end());
        int res = INT_MAX;
        for(int i = 0; i < time.size() - 1; i ++ ) {
            res = min(res, time[i + 1] - time[i]);
        }
        res = min(res, (24 * 60) - time[time.size() - 1] + time[0]);
        return res;
    }
};

更简便的写法：

class Solution {
public:
    int findMinDifference(vector<string>& timePoints) {
        int res = INT_MAX;
        vector<int> q;
        for(auto t : timePoints) {
            int h, m;
            sscanf(t.c_str(), "%d:%d", &h, &m);
            q.push_back(h * 60 + m);
        }
        sort(q.begin(), q.end());
        for(int i = 1; i < q.size(); i ++ ) res = min(res, q[i] - q[i - 1]);
        res = min(res, 24 * 60 - q.back() + q[0]);
        return res;
    }
};