[算法分析与设计] leetcode 每周一题: Minimum Time Difference

题目链接：https://leetcode.com/problems/minimum-time-difference/description/

题目：

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:

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.

思路：

实际上对整个数组排序，然后相邻两两对比，取差值最小的返回（注意：最后一个和第一个的差值也要对比）

代码如下：

class Solution {
public:
    int difference(string a0, string b0) {
        int a1 = (a0[0] - '0') * 600 + (a0[1] - '0') * 60 + (a0[3] - '0') * 10 + (a0[4] - '0');
        int b1 = (b0[0] - '0') * 600 + (b0[1] - '0') * 60 + (b0[3] - '0') * 10 + (b0[4] - '0');
        int result =  abs(a1 - b1);
        return result > 1440 - result ? 1440 - result : result;
    }

    int findMinDifference(vector<string>& timePoints) {
        sort(timePoints.begin(), timePoints.end());
        int result = INT_MAX;
        int len = timePoints.size();
        for(int i = 0; i < len; i++ ) {
            int diff = difference(timePoints[i], timePoints[(i + 1)%len]);
            result = result > diff ? diff : result;
        }

        

        return result;
    }
};