LeetCode 436. Find Right Interval

原题链接在这里：https://leetcode.com/problems/find-right-interval/

题目：

Given a set of intervals, for each of the interval i, check if there exists an interval j whose start point is bigger than or equal to the end point of the interval i, which can be called that j is on the "right" of i.

For any interval i, you need to store the minimum interval j's index, which means that the interval j has the minimum start point to build the "right" relationship for interval i. If the interval j doesn't exist, store -1 for the interval i. Finally, you need output the stored value of each interval as an array.

Note:

1. You may assume the interval's end point is always bigger than its start point.
2. You may assume none of these intervals have the same start point.

Example 1:

Input: [ [1,2] ]

Output: [-1]

Explanation: There is only one interval in the collection, so it outputs -1.

Example 2:

Input: [ [3,4], [2,3], [1,2] ]

Output: [-1, 0, 1]

Explanation: There is no satisfied "right" interval for [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point;
For [1,2], the interval [2,3] has minimum-"right" start point.

Example 3:

Input: [ [1,4], [2,3], [3,4] ]

Output: [-1, 2, -1]

Explanation: There is no satisfied "right" interval for [1,4] and [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point.

NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

题解：

Use a TreeMap tm to stroe mappings between interval start and index i.

For each interval, find ceiling key on interval's end. That would be the first right interval.

Time Complexity: O(nlogn). n = intervals.length.

Space: O(n).

AC Java: 

 1 class Solution {
 2     public int[] findRightInterval(int[][] intervals) {
 3         if(intervals == null || intervals.length == 0){
 4             return new int[0];
 5         }
 6         
 7         int n = intervals.length;
 8         int [] res = new int[n];
 9         TreeMap<Integer, Integer> tm = new TreeMap<>();
10         for(int i = 0; i<n; i++){
11             tm.put(intervals[i][0], i);
12         }
13         
14         for(int i = 0; i<n; i++){
15             Integer key = tm.ceilingKey(intervals[i][1]);
16             if(key == null){
17                 res[i] = -1;
18             }else{
19                 res[i] = tm.get(key);
20             }
21         }
22         
23         return res;
24     }
25 }