Pulling drills - 56 Finding the right interval

56 寻找右区间

1.问题描述
给定一组区间（Contains the starting point and end point）,对于每一个区间 i,检查是否存在一个区间 j,它的起始点大于或等于区间 i 的终点,这可以称为 j 在 i 的“右侧”.

对于任何区间,你需要存储的满足条件的区间 j 的最小索引,这意味着区间 j 有最小的起始点可以使其成为“右侧”区间.如果区间 j 不存在,则将区间 i 存储为 -1.最后,你需要输出一个值为存储的区间值的数组.

注意:

你可以假设区间的终点总是大于它的起始点.

你可以假定这些区间都不具有相同的起始点.

示例 1:

输入: [ [1,2] ]

输出: [-1]

解释:集合中只有一个区间,所以输出-1.

示例 2:

输入: [ [3,4], [2,3], [1,2] ]

输出: [-1, 0, 1]

解释:对于[3,4],没有满足条件的“右侧”区间.

对于[2,3],区间[3,4]具有最小的“右”起点;

对于[1,2],区间[2,3]具有最小的“右”起点.

示例 3:

输入: [ [1,4], [2,3], [3,4] ]

输出: [-1, 2, -1]

解释:对于区间[1,4]和[3,4],没有满足条件的“右侧”区间.

对于[2,3],区间[3,4]有最小的“右”起点.

可使用以下main函数：

int main()

{

vector<vector<int> > intervals;



int n,start,end;

cin>>n;



for(int j=0; j<n; j++)

{

    vector<int> aInterval;

    cin>>start>>end;

    aInterval.push_back(start);

    aInterval.push_back(end);

    intervals.push_back(aInterval);

}

vector<int> res=Solution().findRightInterval(intervals);

for(int i=0; i<res.size(); i++)

{

    if (i>0)

        cout<<" ";

    cout<<res[i];

}

return 0;

}
2.输入说明
The first input interval numbern,

然后输入n行,每行包括两个整数,表示起点和终点
3.输出说明
The contents of the output array,每两个元素之间以一个空格分隔.
4.范例
输入
3
1 4
2 3
3 4
输出
-1 2 -1
5.代码

#include <iostream>
#include <queue>
#include <cstdlib>
#include <cstring>
#include<algorithm>

using namespace std;


vector<int> findRightInterval(vector<vector<int> >intervals) {
    
	
	/*法一：Part of the sample pass,why?*/
	/* //Record each interval the left endpoint locationi vector<pair<int, int> >Position; int n = intervals.size(); for (int i = 0; i < n; i++) { Position.push_back({ intervals[i][0], i }); } //According to the interval value left smallest sort(intervals.begin(), intervals.end()); vector<int>ans; //遍历,Looking for is bigger than the current right interval endpoint range,And the difference between the smallest that interval sequencei for (int i = 0; i < n; i++) { int num = -1; int left = 0; int right = n-1 ; int cur = intervals[i][1];//当前区间右端点 while (left < right) { int mid = (left + right) / 2; if (intervals[mid][0] >=cur) right = mid ;//左区间 else left = mid+1 ;//右区间 } if (intervals[right][0]>=cur) { //cout << "left=" << left << endl; for (int t = 0; t < n; t++) { if (Position[t].first == intervals[right][0]) num = t; } //num = Position[right].second; } ans.push_back(num); } return ans; */
	//法二：
	//参考题解：https://leetcode.cn/problems/find-right-interval/solution/xun-zhao-you-qu-jian-by-leetcode-solutio-w2ic/

	/*输入： [[3,4][2,3],[1,2]] 输出 [-1,0,1] */
	vector<pair<int, int> > startIntervals;

	int n = intervals.size();
	for (int i = 0; i < n; i++) {
    
		startIntervals.emplace_back(intervals[i][0], i);//Extracting the left endpoint of each interval and the initial subscript to combine,Become a key/value pair
	}
	//此时startIntervals中元素为[{3,0},{2,1},{1,2}]
	sort(startIntervals.begin(), startIntervals.end());//According to each interval of the left endpoint smallest
	//元素变成[{1,2},{2,1},{3,0}]

	vector<int> ans(n, -1);//初始都为-1
	//lower_bound(a,b,val)返回在[a,b）The interval is not less thanval的值
	//注意,This function is to use binary search,区间是左闭右开
	//关于lower_boundFunction use please refer tohttps://blog.csdn.net/qq_38786209/article/details/78470260
	for (int i = 0; i < n; i++) {
    
		auto it = lower_bound(startIntervals.begin(), startIntervals.end(), make_pair(intervals[i][1], 0));//此处itOn behalf of meet the requirements ofstartIntervalsAn element of the array of
		//判断it是否存在,To use the statement below【迭代器用法】
		if (it != startIntervals.end()) {
    //it代表startIntervals中某个键值对{}
			ans[i] = it->second;//it->secondSaid returns the key values of the second element,Which is the original subscript
		}
	}
	return ans;

}

int main()

{
    

	vector<vector<int> > intervals;



	int n, start, end;

	cin >> n;



	for (int j = 0; j < n; j++)

	{
    

		vector<int> aInterval;

		cin >> start >> end;

		aInterval.push_back(start);

		aInterval.push_back(end);

		intervals.push_back(aInterval);

	}

	vector<int> res = findRightInterval(intervals);

	for (int i = 0; i < res.size(); i++)

	{
    

		if (i > 0)

			cout << " ";

		cout << res[i];

	}

	return 0;

}