[leetcode] 436. Find Right Interval

[leetcode] 436. Find Right interval

Description

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.

分析

题目的意思是：给定一个区间的集合，让我们找每个区间的最近右区间，要保证右区间的start要大于等于当前区间的end。

STL的lower_bound函数来找第一个不小于目标值的位置.其他的理解起来应该没啥，用了map函数存放start和i的映射。

代码

/** * Definition for an interval. * struct Interval { * int start; * int end; * Interval() : start(0), end(0) {} * Interval(int s, int e) : start(s), end(e) {} * }; */class Solution {public: vector findRightInterval(vector& intervals) { vector res; map m; for(int i=0; isecond); } return res; }};

参考文献

​​[LeetCode] Find Right Interval 找右区间​​

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