539. Minimum Time Difference

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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.

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Brute force 暴力解法：O(n²), 任意两个时间点求差。效率差，忽略。

排序：O(nlgn)

class Solution {
public:
    int findMinDifference(vector<string>& timePoints) {
        const int SIZE=timePoints.size();
        if(SIZE>1440){
            return 0;
        }
        
        //sorting
        sort(timePoints.begin(),timePoints.end());
        int result=INT_MAX;
        int diff;
        for(int i=0;i<SIZE-1;i++){
            diff=getTimeMins(timePoints[i+1])-getTimeMins(timePoints[i]);
            result = min(result,diff);
            if(result==0){
                return result;
            }
        }
        
        return min(result, getTimeMins(timePoints[0])+1440-getTimeMins(timePoints[SIZE-1]));
    }
private:
    int getTimeMins(string time24Hours){ //time24Hours in the format of "##:##"
        return stoi(time24Hours.substr(0,2))*60+stoi(time24Hours.substr(3));
    }
};

 

优化解法：比O(nlng)排序更好的是linear算法，所有的时间点总共有60*24=1440种可能，可以用数组来存该时间点是否出现过。

class Solution {
public:
    int findMinDifference(vector<string>& timePoints) {
        const int SIZE=1440;
        //there are totally 1440 different time points, if more than this number, there are at least two timepoints are the same
        if(timePoints.size()>SIZE){
            return 0;
        }
        
        vector<bool> mask(SIZE, false);
        for(auto timePoint: timePoints){
            int time=getTimeMins(timePoint);
            if(mask[time]){
                return 0;
            }
            mask[time]=true;
        }
        
        int prev = -1; 
        int first = -1; //the first timepoint
        int last = -1;  //the last timepoint
        int result = INT_MAX;
        for(int i=0;i<SIZE;i++){
            if(mask[i]){
                if(prev!=-1){
                    result = min(result,i-prev);
                }
                prev = i;
                if(first==-1){
                    first=i;
                }
                last = i;
            }
        }
        return min(result,first+1440-last);
    }
private:
    //convert timepoint in string to a number of minutes
    int getTimeMins(string time24Hours){ //time24Hours in the format of "##:##"
        return stoi(time24Hours.substr(0,2))*60+stoi(time24Hours.substr(3));
    }
};