题目 给定一个 24 小时制（小时:分钟）的时间列表，找出列表中任意两个时间的最小时间差并已分钟数表示。 示例 1： 输入: ["23:59","00:00"] 输出: 1 备注: 列表中时间数在 2~20000 之间。 每个时间取值在 00:00~23:59 之间。 C++解法 #include #include #include #include using namespace std; struct Time { int hour = 0; int minute = 0; Time(string value = "00:00") { if (value.size() == 5) { for (int i = 0; i < 2; ++i) { hour = 10 * hour + (value[i] - '0'); } for (int i = 3; i < 5; ++i) { minute = 10 * minute + (value[i] - '0'); } } } int gap(Time & time) { int value1 = hour * 60 + minute; int value2 = time.hour * 60 + time.minute; int minValue = value1 < value2 ? value1 : value2; int maxValue = value1 > value2 ? value1 : value2; return min(maxValue - minValue, 1440 - maxValue + minValue); } }; class Solution { public: int findMinDifference(vector& timePoints) { sort(timePoints.begin(), timePoints.end()); string item; int minValue = INT32_MAX; int delta = 0; Time lastTime, currentTime; for (int i = 0; i <= timePoints.size(); ++i) { currentTime = Time(timePoints[i % timePoints.size()]); if (i != 0) { delta = currentTime.gap(lastTime); if (delta < minValue) minValue = delta; } lastTime = currentTime; } return minValue; } }; int main(int argc, const char * argv[]) { // insert code here... Solution solution; vector times; times = {"00:00", "23:59"}; cout << solution.findMinDifference(times) << endl; return 0; } 来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/minimum-time-difference
