LeetCode 539. Minimum Time Difference

原题链接在这里：https://leetcode.com/problems/minimum-time-difference/description/

题目：

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.

题解：

利用bucket sort. 把时间转化成的位置标记成true. 若是之前已经标记了说明有duplicate, 可以直接return 0.

从头到尾iterate buckets维护最小diff. 再和首尾相差的diff比较取出最小值.

Time Complexity: O(timePoints.size() + 24*60).

Space: O(24*60).

AC Java:

class Solution {
    public int findMinDifference(List<String> timePoints) {
        boolean [] mark = new boolean[24*60];
        for(String s : timePoints){
            String [] parts = s.split(":");
            int time = Integer.valueOf(parts[0]) * 60 + Integer.valueOf(parts[1]);
            
            if(mark[time]){
                return 0;
            }
            
            mark[time] = true;
        }
        
        int res = 24*60;
        int pre = -1;
        int first = Integer.MAX_VALUE;
        int last = Integer.MIN_VALUE;
        for(int i = 0; i<24*60; i++){
            if(mark[i]){
                if(pre != -1){
                    res = Math.min(res, i-pre);
                }
                
                pre = i;
                first = Math.min(first, i);
                last = Math.max(last, i);
            }
        }
        
        res = Math.min(res, first+24*60-last);
        return res;
    }
}