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题目描述:
- 给定一个 24 小时制(小时:分钟 "HH:MM")的时间列表,找出列表中任意两个时间的最小时间差并以分钟数表示。
-  
- 示例 1:
- 输入:timePoints = ["23:59","00:00"]
- 输出:1
- 示例 2:
- 输入:timePoints = ["00:00","23:59","00:00"]
- 输出:0
-  
- 提示:
- 2 <= timePoints <= 2 * 104
- timePoints[i] 格式为 "HH:MM"
- 来源:力扣(LeetCode)
- 链接:https://leetcode-cn.com/problems/minimum-time-difference
- 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
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- class Solution {
- public:
- int findMinDifference(vector<string>& timePoints) {
- //转化成分钟
- vector<int>store;
- int len = timePoints.size();
- for(int i = 0; i < len; i++){
- string temp = timePoints[i];
- int temp1 = stoi(temp.substr(0, 2))*60 + stoi(temp.substr(3, 2));
- store.push_back(temp1);
- }
- sort(store.begin(), store.end());
- int len1 = store.size();
- int res = min(store[len1-1] - store[0], store[0]+ 24*60- store[len1 - 1]);//注意这一点!!!
- for(int i = 1 ; i < len1; i++){
- res = min(store[i] - store[i-1], res);
- }
- return res;
- }
- };
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