[leetcode] 539. Minimum Time Difference

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本文介绍了一种计算多个24小时制时间点间最小分钟差的方法。通过排序和比较相邻时间点来找到最短时间间隔，特别关注了跨零点的情况。

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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:

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

Note:

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

这道题是计算多个时间点之间的最小时间间隔，题目难度为Medium。

首先将所有时间点排序，依次比较相邻两个时间点即可，这里需要注意首尾两个时间点也需要比较，另外还需要注意示例中过零点的情况。字符串处理这里用stoi函数来解决。具体代码：

class Solution {
public:
    int findMinDifference(vector<string>& timePoints) {
        int minDiff = INT_MAX;
        sort(timePoints.begin(), timePoints.end());
        for(int i=0; i<timePoints.size(); ++i) {
            int cur = i, pre = i ? i-1 : timePoints.size()-1;
            int curHr = stoi(timePoints[cur].substr(0, 2));
            int preHr = stoi(timePoints[pre].substr(0, 2));
            int curMt = stoi(timePoints[cur].substr(3, 2));
            int preMt = stoi(timePoints[pre].substr(3, 2));
            if(curHr < preHr || (curHr == preHr && curMt < preMt)) curHr += 24;
            minDiff = min(minDiff, (curHr-preHr)*60+curMt-preMt);
        }
        return minDiff;
    }
};

另外还可以采用桶排序的思路来解决问题，拿空间换时间，时间复杂度有所提高，空间复杂度有所降低，感兴趣的同学可以自己实现下。