Find Right Interval

最新推荐文章于 2021-12-28 00:23:53 发布

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最新推荐文章于 2021-12-28 00:23:53 发布

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本文介绍了一种使用哈希映射解决区间匹配问题的方法。对于给定的一组区间，我们需要找到每个区间右侧最近的一个区间，并返回这些区间的索引值。通过构建哈希映射，能够有效地追踪每个区间的起始点及其对应的索引。

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.

感觉用了非主流的做法，击败人数比较低..讲一讲我的思路，我是采用两个hashmap来解这道题的。题目的要求是，求出当前区间紧挨着的右区间的index值，如果不存在那么设置成-1。我的想法是，先用一个ha shmap保存原始数组中interval[i].start跟i （也就是start的值与当前元素下标的对应关系的ha sh ma p）。然后，再复制出一个按照start升序排列的新数组。然后再遍历的过程中，内层循环用于寻找距离它最近靠右的那个区间，保存到另个ha shmap中，这个新的hashmap保存的是当前interval[i].start与跟他