leetcode 436. Find Right Interval 最右边的区间

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.

题意很简单，就是找到最靠近自己的右边的区间

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>#include <cmath>using namespace std;/*struct Interval {     int start;     int end;     Interval() : start(0), end(0) {}     Interval(int s, int e) : start(s), end(e) {}};*/class Solution {public:    vector<int> findRightInterval(vector<Interval>& intervals)     {        map<int, int> hash;        vector<int> res;        int n = intervals.size();        for (int i = 0; i < n; ++i)            hash[intervals[i].start] = i;        for (Interval in : intervals)        {            map<int,int>::iterator itr = hash.lower_bound(in.end);            if (itr == hash.end())                 res.push_back(-1);            else                 res.push_back(itr->second);        }        return res;    }};