406. Queue Reconstruction by Height

Description

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]]

Solution

Insert sort

要么零分要么满分的一道题，实在是很精妙啊。精髓之处是先把个子最高的人挑出来排个序，再把次高的人挑出来插进去，这时最高人的k值不会改变！

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.
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]]
…

class Solution {
    public int[][] reconstructQueue(int[][] people) {
        Arrays.sort(people, new Comparator<int[]>() {
            public int compare(int[] a, int[] b) {
                return a[0] == b[0] ? a[1] - b[1] : b[0] - a[0];
            } 
        });
        
        List<int[]> queue = new LinkedList<>();
        for (int[] curr : people) {
            queue.add(curr[1], curr);   // so elegant!
        }
        
        return queue.toArray(new int[people.length][]);
    }
}