Queue Reconstruction by Height

标签：algorithm   ESS   after   esc   random   style   ons   linked   pos   

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]]

分析：https://leetcode.com/problems/queue-reconstruction-by-height/discuss/89345/Easy-concept-with-PythonC%2B%2BJava-Solution

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.

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]]

 1 public class Solution {
 2     public int[][] reconstructQueue(int[][] people) {
 3         Arrays.sort(people, (a, b) -> a[0] == b[0] ? a[1] - b[1] : b[0] - a[0]);
 4         List<int[]> list = new LinkedList<>();
 5         for (int[] p : people) {
 6             list.add(p[1], p);
 7         }
 8         return list.toArray(new int[list.size()][]);
 9     }
10 }

Queue Reconstruction by Height

标签：algorithm   ESS   after   esc   random   style   ons   linked   pos   

原文地址：https://www.cnblogs.com/beiyeqingteng/p/10992934.html