LeetCode 406. Queue Reconstruction by Height

原题链接在这里：https://leetcode.com/problems/queue-reconstruction-by-height/description/

题目：

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. 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]]

题解：

先把最高的人挑出来，然后按照k把他们插入到ArrayList<int []> resList中. 然后再插第二高的，以此类推.

Time Complexity: O(nlogn). n = people.length. sort用时O(nlogn), 插入时因为使用ArrayList, 可以忽略resize的用时, 所以插入共用时O(n).

Space: O(n).

AC Java:

 1 class Solution {
 2     public int[][] reconstructQueue(int[][] people) {
 3         if(people == null || people.length == 0|| people[0].length != 2){
 4             return people;
 5         }
 6         
 7         Arrays.sort(people, new Comparator<int []>(){
 8             public int compare(int [] a, int [] b){
 9                 if(a[0] == b[0]){
10                     return a[1]-b[1];
11                 }
12                 return b[0]-a[0];
13             }
14         });
15         
16         ArrayList<int []> resList = new ArrayList<int []>();
17         for(int [] item : people){
18             resList.add(item[1], item);
19         }
20         
21         return resList.toArray(new int[people.length][2]);
22     }
23 }