436. Find Right Interval**

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.

    public int[] findRightInterval(Interval[] intervals) {        TreeMap<Integer, Integer> map = new TreeMap<>();        int[] res = new int[intervals.length];        for(int i=0; i<intervals.length;i++){            map.put(intervals[i].start, i);        }        for(int i=0;i<intervals.length;i++){            Integer key = map.ceilingKey(intervals[i].end);            res[i]=key!=null?map.get(key):-1;        }        return res;    }

总结：好聪明，关键在于没有相同的start point。