You are given an array of intervals, where intervals[i] = [starti, endi] and each starti is unique.
The rightinterval for an interval i is an interval j such that startj``>= endi and startj is minimized.
Return an array of right interval indices for each interval i. If no right interval exists for interval i, then put -1 at index i.
Example 1:
**Input:** intervals = [[1,2]]
**Output:** [-1]
**Explanation:** There is only one interval in the collection, so it outputs -1.
Example 2:
**Input:** intervals = [[3,4],[2,3],[1,2]]
**Output:** [-1,0,1]
**Explanation:** There is no right interval for [3,4].
The right interval for [2,3] is [3,4] since start0 = 3 is the smallest start that is >= end1 = 3.
The right interval for [1,2] is [2,3] since start1 = 2 is the smallest start that is >= end2 = 2.
Example 3:
**Input:** intervals = [[1,4],[2,3],[3,4]]
**Output:** [-1,2,-1]
**Explanation:** There is no right interval for [1,4] and [3,4].
The right interval for [2,3] is [3,4] since start2 = 3 is the smallest start that is >= end1 = 3.
Constraints:
1 <= intervals.length <= 2 * 104
intervals[i].length == 2
-106 <= starti <= endi <= 106
The start point of each interval is unique.
ac
classSolution {publicint[] findRightInterval(Interval[] intervals) {// edge casesif (intervals ==null||intervals.length==0) returnnewint[0];int n =intervals.length;TreeMap<Integer,Integer> tmap =newTreeMap<>();for (int i =0; i < n ; i++) {if (!tmap.containsKey(intervals[i].start)) {tmap.put(intervals[i].start, i); } }int[] res =newint[n];for (int i =0; i < n; i++) {Integer ceil =tmap.ceilingKey(intervals[i].end);if (ceil ==null) { res[i] =-1; } else { res[i] =tmap.get(ceil); } }return res; }}/*1) TreeMap record every start val end its first index; 2) iterate, find ceiling of start, get the val*/