力扣-539. 最小时间差

FGGIT

发布于 2024-10-15 15:17:14

5030

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。

class Solution {
    int getMinutes(string &t) {
        return (int(t[0] - '0') * 10 + int(t[1] - '0')) * 60 + int(t[3] - '0') * 10 + int(t[4] - '0');
    }

public:
    int findMinDifference(vector<string> &timePoints) {
        sort(timePoints.begin(), timePoints.end());
        int ans = INT_MAX;
        int t0Minutes = getMinutes(timePoints[0]);
        int preMinutes = t0Minutes;
        for (int i = 1; i < timePoints.size(); ++i) {
            int minutes = getMinutes(timePoints[i]);
            ans = min(ans, minutes - preMinutes); // 相邻时间的时间差
            preMinutes = minutes;
        }
        ans = min(ans, t0Minutes + 1440 - preMinutes); // 首尾时间的时间差
        return ans;
    }
};

方法：