39. 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.

求数组中，两点之间时钟间隔最小值。

方法一：先将时间表达式转化为分钟整数，排序，依次判断相邻两点的分钟间隔，注意跨零点的情况

方法二：因为时间表达式转化为分钟后，整数最大值为23*60+59，该数值不是很大，因此可以采用桶排序，程序如下所示：

class Solution {
    private final int MAX_LEN = 24*60;
    public int findMinDifference(List<String> timePoints) {
        int len = timePoints.size();
        boolean[] bucket = new boolean[MAX_LEN];
        int hour = 0, minute = 0, cur = 0;
        int left = MAX_LEN, right = 0;
        for (String val : timePoints){
            hour = Integer.parseInt(val.substring(0, 2));
            minute = Integer.parseInt(val.substring(val.length() - 2, val.length()));
            cur = hour * 60 + minute;
            if (bucket[cur]){
                return 0;
            }
            bucket[cur] = true;
            left = Math.min(left, cur);
            right = Math.max(right, cur);
        }
        int res = MAX_LEN;
        Integer pre = null;
        while (left <= right){
            while (!bucket[left]){
                left ++;
            }
            if (pre != null){
                res = Math.min(res, Math.min(left - pre, pre + MAX_LEN - right));
            }
            pre = left;
            left ++;
        }
        return res;
    }
}