406. Queue Reconstruction by Height

#### QUESTION:

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]]
``````

#### EXPLANATION:

1.首先是将数组进行排序，将数组由大到小排序，如果数字相同那么就将数字小的排在前面，那么第一轮就可以确定最大数的位置。

2.接着就可以继续第二大数的排序，比如第二个数是6.1，那么就将6插在1的位置上，同时将其他位数后推。

#### SOLUTION:

``````public class Solution {
public int[][] reconstructQueue(int[][] people) {
Arrays.sort(people, (o1, o2) -> {
if(o1==o2)
return o1-o2;
return o2-o1;
});

int n = people.length;
ArrayList<int[]> tmp = new ArrayList<>();
for (int i = 0; i < n; i++)

int[][] res = new int[people.length];
int i = 0;
for (int[] k : tmp) {
res[i] = k;
res[i++] = k;
}

return res;
}
}
``````
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