Queue Reconstruction by Height问题及解法_queue reconstruction problem-优快云博客

本文介绍了一种算法，用于根据人员的高度及其前面特定高度的人数来重构队列。该算法首先挑选出最高的一组人并排序，然后按次高组依次插入，确保最终队列符合给定条件。

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问题描述：

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.

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]]

问题分析（这里省事了，直接摘抄英文~）：

1.Pick out tallest group of people and sort them in a subarray (S). Since there's no other groups of people taller than them, therefore each guy's index will be just as same as his k value.

2.For 2nd tallest group (and the rest), insert each one of them into (S) by k value. So on and so forth.

过程详见代码：

class Solution {
public:
    vector<pair<int, int>> reconstructQueue(vector<pair<int, int>>& people) {
        auto com = [](const pair<int, int>& p1, const pair<int, int>& p2)
		{return p1.first > p2.first || (p1.first == p2.first && p1.second < p2.second);};

		sort(people.begin(), people.end(), com);
		vector<pair<int, int>> res;
		for (auto p : people)
		{
			res.insert(res.begin() + p.second, p);
		}
		return res;
    }
};