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.
NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
寻找右区间。
题意是给一个list，里面装的是一些interval区间，对于每个interval i，请你寻找是否存在另外一个interval j，满足j在i的右边。在右边的定义是j的起点要大于等于i的终点。对于每一个interval i，如果能找到一个这样的j区间，则返回j的下标，否则返回-1。
这个题二分法可以做，但是代码不是很好写，面试的时候如果不熟练很容易写不完。我这里介绍一个treemap的做法。遍历input，将每个区间的起点当做key，每个区间的下标当做value放入treemap。再次遍历input，此时会利用到treemap的一个函数叫做ceilingKey()，找当前map中是否存在一个最小的区间的起点，这个起点比当前遍历到的区间的终点要大。如果找到，则放入结果集。
时间O(nlogn)
空间O(n)
Java实现
1 class Solution {
2 public int[] findRightInterval(int[][] intervals) {
3 int[] res = new int[intervals.length];
4 TreeMap<Integer, Integer> map = new TreeMap<>();
5 for (int i = 0; i < intervals.length; i++) {
6 map.put(intervals[i][0], i);
7 }
8
9 for (int i = 0; i < intervals.length; i++) {
10 Integer key = map.ceilingKey(intervals[i][1]);
11 res[i] = key != null ? map.get(key) : -1;
12 }
13 return res;
14 }
15 }