Minimum Time Difference问题及解法

问题描述：

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.

示例:

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.

问题分析：

将时间转换为整数值，然后升序排序，计算相邻数值之间的差值（最后一个和第一个算做系相邻元素），最小的差值即为答案。

过程详见代码：

class Solution {public:    int findMinDifference(vector<string>& timePoints) {        sort(timePoints.begin(), timePoints.end());int mini = INT_MAX;int hour1 = stoi((timePoints[0]).substr(0, 2)),hour2;int minute1 = stoi((timePoints[0]).substr(3, 2)),minute2;for (int i = 1; i < timePoints.size(); i++){hour2 = stoi((timePoints[i]).substr(0, 2));minute2 = stoi((timePoints[i]).substr(3, 2));mini = min(mini, (hour2 - hour1) * 60 + (minute2 - minute1));hour1 = hour2;minute1 = minute2;}int m = (23 - hour1) * 60 + (60 - minute1) ;hour1 = stoi((timePoints[0]).substr(0, 2));minute1 = stoi((timePoints[0]).substr(3, 2));m = m + hour1 * 60 + minute1;mini = min(mini, m);return mini;    }};