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:
You may assume the interval's end point is always bigger than its start point.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. 首先，这道题让我们按照每个间隔的右点，遍历间隔数组，寻找大于或等于的最接近的间隔的左节点。那么我们需要一个map存左节点和index，这里用treemap保存自动排序，并用ceilingkey寻找lowbound的key，使代码极为简洁。代码如下： /** * Definition for an interval. * public class Interval { * int start; * int end; * Interval() { start = 0; end = 0; } * Interval(int s, int e) { start = s; end = e; } * } */ public class Solution { public int[] findRightInterval(Interval[] intervals) { TreeMap<Integer, Integer> map = new TreeMap<Integer, Integer>(); int[] res = new int[intervals.length]; for (int i = 0; i < intervals.length; i ++) { map.put(intervals[i].start, i); } for (int j = 0; j < intervals.length; j ++) { Integer key = map.ceilingKey(intervals[j].end); res[j] = key == null? -1: map.get(key); } return res; } }