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]]
题目描述:
一直一串数组,对数组进行排序。每个数组是(h,k)的形式,h代表数组值的大小,k代表前面有几个大于或者等于该数组值的个数。
解题思路:
首先将数组按照h降序 ,k升序进行排序。即:先比较h值的大小(按照h降序排序),如果h值相同,比较k值(按照k值升序进行排序),然后从头向尾遍历数组串,将每个数组放在队列k的位置。具体代码及思路如下:
package Leetcode_Github; import java.util.ArrayList; import java.util.Arrays; import java.util.List; public class GreedyThought_ReconstructQueue_406_1108 { public int[][] reconstructQueue(int[][] people){ if (people == null || people.length == 0 || people[0].length == 0) { return new int[0][0]; } //按照身高降序,K值升序进行排序 Arrays.sort(people, (m, n) -> (m[0] == n[0] ? m[1] - n[1] : n[0] - m[0])); //将people中的每个数组放到队列peo[1]处 List<int[]> queue = new ArrayList<>(); for (int[] peo : people) { queue.add(peo[1], peo); } //将队列转换成数组,返回 return queue.toArray(new int[queue.size()][]); } }