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 当 index 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(n^2). n = people.length. sort用时O(nlogn), ArrayList add(index, val) takes O(n) time, and there are n items.

Space: O(n).

AC Java:

class Solution {
    public int[][] reconstructQueue(int[][] people) {
        if(people == null || people.length == 0){
            return people;
        }
        
        Arrays.sort(people, (a, b) -> a[0] == b[0] ? a[1] - b[1] : b[0] - a[0]);
        ArrayList<int []> res = new ArrayList<>();
        for(int [] p : people){
            res.add(p[1], p);
        }
        
        return res.toArray(new int[people.length][2]);
    }
}

跟上Count of Smaller Numbers After Self.