Queue Reconstruction by Height(算法分析week8)

Queue Reconstruction by Height

题目来源：https://leetcode.com/problemset/algorithms/
题目类型：贪心算法

-题目描述-
-解题思路-
-代码实现-

题目描述

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]]

解题思路

题意为将队列按照(h, k)重新排列，(h, k)用来描述一个人的身高和他前面比他高的人的数目。不断将满足k=0的人中身高最低的添加到结果集中可以实现将队列按照(h, k)重新排列。
(1)将队列中k=0的人中身高最低(hi,ki)添加到结果集中，并从队列中删除这个人。将剩下的人中身高小于等于hi的人的k值减-1。
(2)重复步骤一，直到队列为空。

代码实现

@requires_authorizationclass Solution {public:    vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {        vector<pair<int, int>> temp = people;        vector<pair<int, int>> result;        int j = 0;        while (!people.empty()) {            int min = INT_MAX;            int pos = 0;            for (int i = 0; i < people.size(); i++) {                if (people[i].second == 0&&people[i].first < min) {                    min = people[i].first;                    pos = i;                }            }            int h = people[pos].first;            for (int i = 0; i < result.size(); i++) {                if (h <= result[i].first) people[pos].second++;            }            result.push_back(make_pair(people[pos].first, people[pos].second));            people.erase(people.begin() + pos);            for (int i = 0; i < people.size(); i++) {                if (people[i].first <= h) people[i].second -= 1;            }            j++;        }        return result;    }};