# 【LeetCode】 406. Queue Reconstruction by Height

## 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.

> 假設有隨機的人群在列隊中。每個人有一對整數`(h, k)`去描述他，`h`是這個人的身高，`k`是他前面有多少人比他高。寫一個演算法重新排序這個列隊。

> 提示：
> 描述人的數字小於 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]]

## Solution

* 先讓人按照身高遞減排序。
* 接著從頭跑一次列隊再重新排序一次，這次他前面有幾個人比他高就將他安插在哪個位子。
* 因為列隊已經按照身高排序，後面的一定都比較矮，不會因為後面的人加入導致前面的人不符合`k`的規則。

### Code

bool myCompare(vector<int> a, vector<int> b)
{
    return (a[0] > b[0] || (a[0] == b[0] && a[1] < b[1]));
}

class Solution {
public:
    vector<vector<int>> reconstructQueue(vector<vector<int>>& people) {
        vector<vector<int>> ans;
        sort(people.begin(), people.end(), myCompare);
        for(int i = 0; i < people.size(); i++)
            ans.insert(ans.begin() + people[i][1], people[i]);
        return ans;
    }
};

###### tags: `LeetCode` `C++`