LeetCode – Queue Reconstruction by Height (Java)

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]]

Analysis

The key to solve this problem is finding the start point of reconstruction. For this problem, we can start adding the largest element first. The basic idea is to keep adding the largest element each time, until all elements are in place.

Java Solution

public int[][] reconstructQueue(int[][] people) {
    int[][] result = new int[people.length][];
    Arrays.sort(people, new Comparator<int[]>(){
        public int compare(int[] a1, int[] a2){
            if(a1[0]!=a2[0]){
                return a2[0]-a1[0];
            }else{
                return a1[1]-a2[1];
            }
        }
    });
 
    ArrayList<int[]> list = new ArrayList<int[]>();
 
    for(int i=0; i<people.length; i++){
        int[] arr = people[i];
        list.add(arr[1],arr);
    }
 
    for(int i=0; i<people.length; i++){
        result[i]=list.get(i);
    }
 
    return result;
}

Time complexity of this solution is O(n^2).