LeetCode 406. Queue Reconstruction by Height

406. Queue Reconstruction by Height

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  • Difficulty: Medium
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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]]

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【题目分析】

题目给出了一些人的身高和这个人前面身高大于等于他的人数,对这些人进行排队,使得所有人的情况都得到满足。

【思路】

  1. 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.
  2. For 2nd tallest group (and the rest), insert each one of them into (S) by k value. So on and so forth.

首先找到身高最高的人并对他们进行排序。

然后找到身高次高的人,按照他们的前面的人数把他们插入到最高的人群中。

因此这是一个排序和插入的过程,按照身高进行降序排序,然后把身高相同的人按照k进行升序排序。每次取出身高相同的一组人,按照k值把他们插入到队列中。

【java代码】

 1 public class Solution {
 2     public int[][] reconstructQueue(int[][] people) {
 3         Arrays.sort(people, new Comparator<int[]>(){
 4             public int compare(int[] a, int[] b) {
 5                 if(a[0] != b[0]) return -a[0]+b[0];
 6                 else return a[1]-b[1];
 7             }
 8         });
 9         
10         List<int[]> res = new LinkedList<>();
11         for(int[] p : people) {
12             res.add(p[1], p);
13         }
14         
15         return res.toArray(new int[people.length][]);
16     }
17 }

 

posted @ 2017-02-15 20:56  Black_Knight  阅读(655)  评论(0编辑  收藏  举报