436. Find Right Interval_436. 寻找右区间-优快云博客

本文链接：https://blog.youkuaiyun.com/zshouyi/article/details/71330544

本文介绍了一种算法，用于解决给定一系列区间后，如何找出对于每个区间i是否存在另一个区间j，使得j的起始点大于等于i的结束点的问题。通过使用TreeMap，实现了高效的查找最接近的符合条件的区间j，并返回所有区间的索引结果。

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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;
    }
}