406. 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]]

非常好理解的思路：

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.
For 2nd tallest group (and the rest), insert each one of them into (S) by k value. So on and so forth.
E.g.
input: [[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]
subarray after step 1: [[7,0], [7,1]]
subarray after step 2: [[7,0], [6,1], [7,1]]

代码如下：

public class Solution {    public int[][] reconstructQueue(int[][] people) {        Arrays.sort(people, (a, b) -> a[0] == b[0] ? a[1] - b[1] : b[0] - a[0]);        LinkedList<int[]> list = new LinkedList<int[]>();        for (int[] p: people) {            list.add(p[1], p);        }        return list.toArray(new int[list.size()][]);    }}

按下面的方法，重写compare函数速度会比较快。代码如下：

public class Solution {    public int[][] reconstructQueue(int[][] people) {        Arrays.sort(people,new Comparator<int[]>(){           @Override           public int compare(int[] o1, int[] o2){               return o1[0]!=o2[0]?-o1[0]+o2[0]:o1[1]-o2[1];           }        });        List<int[]> res = new LinkedList<>();        for(int[] cur : people){            res.add(cur[1],cur);               }        return res.toArray(new int[people.length][]);    }}