# Queue Reconstruction by Height

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]]
```

分析：

先按身高倒序，index正序排序，override comparator class's function compare。然后每次按照index插入array。当前已有的元素则后退。最后用 ret.toArray(new int\[0])返回结果。

![](/files/-LjqbD4a4TKuhrgJ1mws)

答案：

```
    public int[][] reconstructQueue(int[][] people) {
        if (people == null || people.length == 0 || people[0].length == 0)
            return new int[0][0];
        int[][] ret = new int[people.length][people[0].length];
        Arrays.sort(people, new Comparator<int[]>(){
            public int compare(int[] a, int[] b){
                if(a[0] == b[0]){
                    return a[1] - b[1];
                }
                return b[0] - a[0];

            }
        });
        List<int[]> temp = new ArrayList<int[]>();
        for(int[] p : people){
            temp.add(p[1], p);//按照Index插入 把上一次该位置的数挤到后面
        }
        return temp.toArray(new int[0][0]);//可以直接用index 0 也可以用实际length
    }
```


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