字符串539. 最小时间差

Posted 2020-11-27 ocpc

题目：

 

 

 

 

解答：

思路：

1，时间转化为分钟数；

2，然后对数字进行排序，进行比较；

3，注意头部和尾部时间的比较时需要考虑不同的方向；

 1 class Solution {
 2 public:
 3     const int DAY_MINUTE = 24 * 60;
 4     int time2int(const string& t) 
 5     {
 6         int hour = stoi(t.substr(0, 2));
 7         int minute = stoi(t.substr(3, 2));
 8         return hour * 60 + minute;
 9     }
10     int findMinDifference(vector<string>& timePoints) 
11     {
12         int N = timePoints.size();
13         vector<int> times(N, 0);
14         for (int i = 0; i < N; ++i) 
15         {
16             times[i] = time2int(timePoints[i]);
17         }
18         sort(times.begin(), times.end());
19         int res = min(times[N - 1] - times[0], times[0] + DAY_MINUTE - times[N - 1]);
20         for (int i = 1; i < N; ++i) 
21         {
22             res = min(res, times[i] - times[i - 1]);
23         }
24         return res;
25     }
26 };