题目描述：

Given a list of 24-hour clock time points in “Hour:Minutes” format, find the minimum minutes difference between any two time points in the list.

Example 1:

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Input: ["23:59","00:00"]
Output: 1

Note:

  1. The number of time points in the given list is at least 2 and won’t exceed 20000.
  2. The input time is legal and ranges from 00:00 to 23:59.

先进行排序，然后依次计算相邻时间的时间差，这个时间差有两个方向，选择较小的一个。最后要注意还要算最后一个时间与第一个时间的时间差。

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class Solution {
public:
    int findMinDifference(vector<string>& timePoints) {
        if (timePoints.size() >= 1440) return 0;
        sort(timePoints.begin(), timePoints.end());
        int minDiff = INT_MAX;
        for (int i = 1; i < timePoints.size(); i++) {
            minDiff = min(minDiff, calcDiff(timePoints[i - 1], timePoints[i]));
        }
        minDiff = min(minDiff, calcDiff(timePoints[0], timePoints.back()));
        return minDiff;
    }
    
    int calcDiff (string &a, string &b) {
        int aH = stoi(a.substr(0, 2)), 
            bH = stoi(b.substr(0, 2)),
            aM = stoi(a.substr(3, 2)),
            bM = stoi(b.substr(3, 2));
            
        int diff;
        
        if (bH == aH) {
            return bM - aM;
        }
        else {
            diff = (60 - aM) + bM + (bH - aH - 1) * 60;
        }
        return min(diff, 1440 - diff);
    }
};