[Google] LeetCode 539 Minimum Time Difference

Given a list of 24-hour clock time points in "HH:MM" format, return the minimum minutes difference between any two time-points in the list.

Solution

这里用到的第一个是 \(stoi\) 函数，将字符串转换成数字。其次为了求最短的时间差，注意的一点是正向和反向：

\[\min(df, 24\cdot 60-df) \]

最后一个细节就是第一个和最后一个的时间差，因为时间是循环的，所以在 \(sort\) 以后，要把第一个时间 \(push\_back\) 进去

点击查看代码

class Solution {
private:
    int dif(string t, string s){
        int ht = stoi(t.substr(0,2));
        int hs = stoi(s.substr(0,2));
        int mt = stoi(t.substr(3,2));
        int ms = stoi(s.substr(3,2));
        int df = abs(60*(ht-hs)+mt-ms);
        return min(df, 1440-df);
    }
public:
    int findMinDifference(vector<string>& timePoints) {
        int n = timePoints.size();
        int ans = INT_MAX;
        sort(timePoints.begin(), timePoints.end());
        timePoints.push_back(timePoints[0]);
        for(int i=1;i<=n;i++){
            ans=min(ans, dif(timePoints[i], timePoints[i-1]));
        }
        return ans;
    }
};