Description

Suppose you have a random list of people standing in a queue. Each person is described by a pair of integers (h, k), where h is the height of the person and k is the number of people in front of this person who have a height greater than or equal to h. Write an algorithm to reconstruct the queue.

Note:

The number of people is less than 1,100.

Example

Input:

[[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]

Output:

[[5,0], [7,0], [5,2], [6,1], [4,4], [7,1]]

#include <iostream>
#include <vector> using namespace std; bool compare(const pair<int, int>& p1, const pair<int, int>& p2)
{
return p1.first > p2.first || (p1.first == p2.first && p1.second < p2.second);
} vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
sort(people.begin(), people.end(), compare);
//按照高度h进行降序排序,如果高度相同就按高于的人数升序排序
//排序后结果[7,0] [7,1] [6,1] [5,0] [5,2] [4,4]
//每一pair都小于等于前面的每一个pair
vector<pair<int, int>> res;
for (auto& p : people)
res.insert(res.begin() + p.second, p);
//根据前面有高于的数目插入相应位置
//如[6,1]插入到位置1,前面只有[7,0],所以满足了[6,1]要求
//[5,0]插入到位置0,满足[5,0]要求
//[5,2]插入到位置2,前面有[5,0] [7,0],满足要求
//……
return res;
} //下面是测试案例
int main()
{ vector<pair<int, int> > pp;
pp.push_back(make_pair<int,int>(7,0));
pp.push_back(make_pair<int,int>(4,4));
pp.push_back(make_pair<int,int>(7,1));
pp.push_back(make_pair<int,int>(5,0));
pp.push_back(make_pair<int,int>(6,1));
pp.push_back(make_pair<int,int>(5,2)); for(auto iter=pp.begin();iter!=pp.end();iter++)
{
cout<<"["<<iter->first<<","<<iter->second<<"]"<<" ";
}
vector<pair<int, int>> res = reconstructQueue(pp);
cout << endl << "-------------------------------" << endl;
for(auto iter=res.begin();iter!=res.end();iter++)
{
cout<<"["<<iter->first<<","<<iter->second<<"]"<<" ";
}
cout << endl;
return 0;
}

输出:

[7,0] [4,4] [7,1] [5,0] [6,1] [5,2]

[5,0] [7,0] [5,2] [6,1] [4,4] [7,1]

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