給定一個 24 小時制（小時:分鐘）的時間列表，找出列表中任意兩個時間的最小時間差並已分鐘數表示。
示例 1：
輸入: ["23:59","00:00"]
輸出: 1
備註:
1.列表中時間數在 2~20000 之間。
2.每個時間取值在 00:00~23:59 之間。
詳見：https://leetcode.com/problems/minimum-time-difference/description/
C++：

class Solution {
public:
    int findMinDifference(vector<string>& timePoints)
    {
        int res = INT_MAX, n = timePoints.size(), diff = 0;
        sort(timePoints.begin(), timePoints.end());
        for (int i = 0; i < n; ++i)
        {
            string t1 = timePoints[i], t2 = timePoints[(i + 1) % n];
            int h1 = (t1[0] - ‘0‘) * 10 + t1[1] - ‘0‘;
            int m1 = (t1[3] - ‘0‘) * 10 + t1[4] - ‘0‘;
            int h2 = (t2[0] - ‘0‘) * 10 + t2[1] - ‘0‘;
            int m2 = (t2[3] - ‘0‘) * 10 + t2[4] - ‘0‘;
            diff = (h2 - h1) * 60 + (m2 - m1);
            if (i == n - 1)
            {
                diff += 24 * 60;
            }
            res = min(res, diff);
        }
        return res;
    }
};

參考：http://www.cnblogs.com/grandyang/p/6568398.html

539 Minimum Time Difference 最小時間差