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:
The number of time points in the given list is at least 2 and won’t exceed 20000.
The input time is legal and ranges from 00:00 to 23:59.

思路：
找相隔最短，这里可以利用bucket法。新建大小24*60的boolean数组，按照h*60+m把所有时间转换成int型，对应的bucket设置为true。在遍历boolean数组，找相隔最短的true。最后要注意比较一下min和24 * 60 - last + first的大小。

class Solution {
    public int findMinDifference(List<String> timePoints) {
        boolean[] times = new boolean[24 * 60];  
        for (String time: timePoints) {  
            String[] t = time.split(":");  
            int h = Integer.parseInt(t[0]);  
            int m = Integer.parseInt(t[1]);  
            if (times[h * 60 + m]) {  
                return 0;  
            }  
            times[h * 60 + m] = true;  
        }  
        int pre = -24 * 60, min = Integer.MAX_VALUE, first = 24 * 60, last = 0;   
        for (int i = 0; i < times.length; i ++) {  
            if (times[i]) {  
                min = Math.min(min, i - pre);  
                if (first >= i) {  
                    first = i;  
                }  
                last = i;  
                pre = i;  
            }  
        }  
        return Math.min(min, 24 * 60 - last + first);  
    }
}