539. Minimum Time Difference

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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

计数排序的思想，线性化序列，最大值最小值已知。

代码

class Solution {
       public int findMinDifference(List<String> timePoints) {
        boolean[] mark = 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 (mark[h * 60 + m]) return 0;
            mark[h * 60 + m] = true;
        }
        
        int prev = 0, min = Integer.MAX_VALUE;
        int first = Integer.MAX_VALUE, last = Integer.MIN_VALUE;
        for (int i = 0; i < 24 * 60; i++) {
            if (mark[i]) {
                if (first != Integer.MAX_VALUE) {
                    min = Math.min(min, i - prev);
                }
                first = Math.min(first, i);
                last = Math.max(last, i);
                prev = i;
            }
        }
        
        min = Math.min(min, (24 * 60 - last + first));
        
        return min;
    }
}