[leetcode-539-Minimum Time Difference]

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.

思路：

将字符串时间转换成唯一的分钟数，映射到0-1439的区间上去。

为了只需要遍历一次，将当前时间t和t+1440都存到map里，这样只需要比较临近的两个时间即可，

比如00:00保存为0和1440，那么23:59保存为1439和1439+1440，当比较到1439和1440的时候，得到最小分钟数为1.

//转换成分钟数
int timeToMinu(string time)
{
    time[2] =' ';
    stringstream ss(time);
    int hour,minute;
    ss>>hour>>minute;
    return hour*60+minute;    
}
int findMinDifference(vector<string>& timePoints)
{
  if(timePoints.size()>1440)return 0;//24小时最大分钟数
  map<int,int>mp;
  for(int i =0;i<timePoints.size();i++)
  {
    if(mp.count( timeToMinu(timePoints[i]) )) return 0;
    mp[timeToMinu(timePoints[i])]++;
    mp[timeToMinu(timePoints[i])+1440]++;
  }
  int ret = 1440;
  auto it1 = mp.begin(),it2 = ++mp.begin();
  for(;it2!=mp.end();it1++,it2++)
  {
    ret = min(ret,it2->first - it1->first);
  }
  return ret;
}