# 1229. Meeting Scheduler
###### tags: `Leetcode` `Medium` `Two Pointers`
Link: https://leetcode.com/problems/meeting-scheduler/
## 思路 $O(NlogN)$ $O(1)$
先根据```arr[0]```排序 然后双指针
注意while的条件 有的时候只要有一个小于length就可以继续跑下去
## Code

class Solution {
    public List<Integer> minAvailableDuration(int[][] slots1, int[][] slots2, int duration) {
        Arrays.sort(slots1, (a,b)->(a[0]-b[0]));
        Arrays.sort(slots2, (a,b)->(a[0]-b[0]));
        int p1=0, p2=0;
        List<Integer> ans = new ArrayList<>();
        while(p1<slots1.length && p2<slots2.length){
            int[] intersect = new int[]{Math.max(slots1[p1][0], slots2[p2][0]), Math.min(slots1[p1][1],slots2[p2][1])};
            if(intersect[0]<intersect[1] && intersect[1]-intersect[0]>=duration){
                ans.add(intersect[0]);
                ans.add(intersect[0]+duration);
                return ans;
            }
            if(slots1[p1][1]>slots2[p2][1]){
                p2++;
            }
            else{
                p1++;
            }
        }
        return ans;
    }
}