Queue Reconstruction by Height

原题链接： https://leetcode.com/problems/queue-reconstruction-by-height/description/

题目： 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.

示例：
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]]

Solution： 这个题我们可以使用如下的算法：首先我们可以给这n个人按身高降序进行排序，那么对于每一个人（h,k）,k就是他对应的在这个队列中的位置，因为比他高以及和他一样高的人都已经放进结果队列中了。例如，示例中，最高的是[7,0]和[7,1]，只需要先将[7,0]按其序号k插入队列即可，得到result = [[7,0]]。再将[7,1]按其k=1插入队列第2位，得到result = [[7,0], [7,1]]。最高的已经放入队列中了，再考虑次高的[6,1]，按其k=1插入队列第2位，得到result = [[7,0], [6,1], [7,1]]。如此类推，直到所有的人都放入队列中。（代码中使用了优先队列来简化操作）

class Solution {
public:
    struct cmp {
      bool operator()(pair<int, int> a, pair<int, int> b) {
          if (a.first == b.first) return a.second > b.second;
          else return a.first < b.first;
      }  
    };
    vectorint, int>> reconstructQueue(vectorint, int>>& people) {
        int n = people.size();
        vectorint, int> > result;
        priority_queueint, int>, vectorint, int> >, cmp> q;

        for (auto p : people) {
            q.push(p);       
        }


        while (!q.empty()) {
            pair<int, int> p = q.top();
            cout << p.first << " " << p.second << endl;
            result.insert(result.begin() + p.second, p);
            q.pop();
        }

        return result;
    }
};