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.
把24小时表示的转成分钟。然后排序。需要注意的是要考虑“23：59”和“00：01”的时间差是2，“00：01”可以相当于“24：01”，所以要考虑最小时间和最大时间的时间差，即在排序玩的vector里，把最小的时间加上1440分钟放到最后。然后一趟循环找到最小的时间。
代码如下：
class Solution {
public:
int findMinDifference(vector<string>& timePoints) {
vector<int> times;
int n = timePoints.size();
for(int i = 0; i < n; i++)
{
times.push_back(str2min(timePoints[i]));
}
sort(times.begin(),times.end());
times.push_back(times[0] + 1440);
int min = times[1] - times[0];
for(int j = 2;j<times.size();j++)
{
if(times[j] - times[j-1] < min)
min = times[j] - times[j-1];
}
return min;
}
private:
//转成分钟数
int str2min(string s)
{
return atoi(s.substr(0, 2).c_str()) * 60 + atoi(s.substr(3, 2).c_str());
}
};
