LeetCode 406. Queue Reconstruction by Height

406. Queue Reconstruction by Height

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• Total Accepted: 16876
• Total Submissions: 30903
• 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】

public class Solution {
    public int[][] reconstructQueue(int[][] people) {
        Arrays.sort(people, new Comparator<int[]>(){
            public int compare(int[] a, int[] b) {
                if(a[0] != b[0]) return -a[0]+b[0];
                else return a[1]-b[1];
            }
        });
        
        List<int[]> res = new LinkedList<>();
        for(int[] p : people) {
            res.add(p[1], p);
        }
        
        return res.toArray(new int[people.length][]);
    }
}