LeetCode #436 - Find Right Interval

最新推荐文章于 2023-01-24 17:24:45 发布

原创最新推荐文章于 2023-01-24 17:24:45 发布 · 150 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 专栏收录该内容

625 篇文章

订阅专栏

本文介绍了一种寻找给定区间集合中每个区间的最近右区间的算法实现。通过构造并排序区间起点数组，利用哈希表建立起点与区间下标的映射关系，最终输出每个区间的最近右区间的下标数组。

题目描述：

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.

给定一组区间，对于每个区间找出对应的最近的右区间，右区间的定义为一个区间j的起点大于等于另一个区间i的终点，那么区间j就是区间i的右区间。这道题需要构造一个区间起点的数组，并对数组排序，这样对于任意区间可以根据它的终点在起点数组搜索最近的数组起点。由于每一个数组的起点都不一样，所以我们可以利用哈希表构造区间起点和下标的对应关系，搜索到了右区间的起点就得到了右区间的下标。

class Solution {
public:
    vector<int> findRightInterval(vector<Interval>& intervals) {
        vector<int> start;
        unordered_map<int,int> hash;
        for(int i=0;i<intervals.size();i++)
        {
            start.push_back(intervals[i].start);
            hash[intervals[i].start]=i;
        }
        sort(start.begin(),start.end());
        
        vector<int> result;
        int max_start=start.back();
        for(int i=0;i<intervals.size();i++)
        {
            if(intervals[i].end>max_start) result.push_back(-1);
            else
            {
                int right_start=*lower_bound(start.begin(),start.end(),intervals[i].end);
                result.push_back(hash[right_start]);
            }
        }
        return result;
    }
};