leetcode1229. Schedule the meeting

subject

You are an administrative assistant , I have two customers' free schedules ：slots1 and slots2, And the expected duration of the meeting duration, Please arrange a suitable meeting time for them .

「 Meeting time 」 Yes, both customers are free to attend , And the duration can meet the expected time duration Of The earliest time interval .

If there is no meeting time to meet the requirements , Please return one An empty array .

「 free time 」 The format is [start, end], From start time start And end time end form , From start Start , To end end .

Ensure that the data is valid ： The same person's free time will not overlap , in other words , For two free times of the same person [start1, end1] and [start2, end2], or start1 > end2, or start2 > end1.

Example 1：

 Input ：slots1 = [[10,50],[60,120],[140,210]], slots2 = [[0,15],[60,70]], duration = 8
 Output ：[60,68]

Example 2：

 Input ：slots1 = [[10,50],[60,120],[140,210]], slots2 = [[0,15],[60,70]], duration = 12
 Output ：[]

Tips ：

1 <= slots1.length, slots2.length <= 10^4
slots1[i].length, slots2[i].length == 2
slots1[i][0] < slots1[i][1]
slots2[i][0] < slots2[i][1]
0 <= slots1[i][j], slots2[i][j] <= 10^9
1 <= duration <= 10^6 

 

Realization

public static int[] free(int[][] nums1, int[][] nums2, int duration) {
        int index1 = 0;
        int index2 = 0;
        int[] result = new int[2];
        while (index1 < nums1.length && index2 < nums2.length) {
            int begin = Math.max(nums1[index1][0], nums2[index2][0]);
            int end = Math.min(nums1[index1][1], nums2[index2][1]);
            if (end - begin >= duration) {
                result[0] = begin;
                result[1] = begin + duration;
                return result;
            } else if (nums2[index2][1] > nums1[index1][1]) {
                index1++;
            } else {
                index2++;
            }
        }

        return result;
    }