題目：

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

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]]

題目描述：

     一直一串陣列，對陣列進行排序。每個陣列是（h,k）的形式，h代表陣列值的大小，k代表前面有幾個大於或者等於該陣列值的個數。

解題思路：

   首先將陣列按照h降序 ，k升序進行排序。即：先比較h值的大小（按照h降序排序），如果h值相同，比較k值（按照k值升序進行排序），然後從頭向尾遍歷陣列串，將每個陣列放在佇列k的位置。具體程式碼及思路如下：

package Leetcode_Github;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class GreedyThought_ReconstructQueue_406_1108 {
    public int[][] reconstructQueue(int[][] people){
        if (people == null || people.length == 0 || people[0].length == 0) {
            return new int[0][0];
        }

        //按照身高降序，K值升序進行排序
        Arrays.sort(people, (m, n) -> (m[0] == n[0] ? m[1] - n[1] : n[0] - m[0]));

        //將people中的每個陣列放到佇列peo[1]處
        List<int[]> queue = new ArrayList<>();
        for (int[] peo : people) {
            queue.add(peo[1], peo);
        }

        //將佇列轉換成陣列，返回
        return queue.toArray(new int[queue.size()][]);
    }
}