LC 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.

比较笨的一种方法就是我的方法，每次转成分钟然后做差就行了，注意要算补数求最小，计算时头和尾也要计算一次差值。

Runtime: 24 ms, faster than 25.52% of C++ online submissions for Minimum Time Difference.

class Solution {
public:
  int minutediff(string time1, string time2){
    int hour1 = atoi(time1.substr(,).c_str());
    int hour2 = atoi(time2.substr(,).c_str());
    int minute1 = atoi(time1.substr(,).c_str());
    int minute2 = atoi(time2.substr(,).c_str());
    minute1 = hour1 *  + minute1;
    minute2 = hour2 *  + minute2;
    int dif1 = abs(minute1 - minute2);
    int dif2 = abs( *  - dif1);
    return min(dif1, dif2);
  }

  int findMinDifference(vector<string>& timePoints) {
    sort(timePoints.begin(), timePoints.end());
    vector<int> timedif(timePoints.size(),INT_MAX);
    for(int i=; i<timePoints.size(); i++){
      timedif[i] = minutediff(timePoints[i],timePoints[i-]);
    }
    timedif[] = minutediff(timePoints[], timePoints.back());
    int ret = *min_element(timedif.begin(),timedif.end());
    return ret;
  }
};

这一种做法时间不快因为每一次时间需要进行两次转换，下面看怎么进行一次转换。

另一种高级的做法，因为总共就24*60种可能，不如开一个24*60的数组，计算差值，妙！

class Solution {
public:
    int findMinDifference(vector<string>& timePoints) {
        vector<int> v( * , );
        int r = INT_MAX, begin =  * , last = - * ;
        for (string & p : timePoints) {
            int t = stoi(p.substr(, )) *   + stoi(p.substr(, ));
            if (v[t] == ) return ;
            v[t] = ;
        }
        for (int i = ; i <  * ; i++) {
            if (v[i]) {
                r = min(r, i - last);
                begin = min(begin, i);
                last = i;
            }
        }
        return min(r, begin +  *  - last);
    }
};