Puzzle: Find the order of n persons standing in a line (based on their heights)

Question

Saw this question on Careercup.com:

Given heights of n persons standing in a line and a list of numbers corresponding to each person (p) that gives the number of pers

Answer 1

This is the implementation for the idea provided by user1990169. Complexity being O(N^2).

public class Solution {
        class Person implements Comparator{
            int height;
            int infront;
            public Person(){

            }
            public Person(int height, int infront){
                this.height = height;
                this.infront = infront;
            }
            public int compare(Person p1, Person p2){
                return p1.height - p2.height;
            }
        }

        public ArrayList order(ArrayList heights, ArrayList infronts) {
           int n = heights.size();
           Person[] people = new Person[n];
           for(int i = 0; i < n; i++){
               people[i] = new Person(heights.get(i), infronts.get(i));
           }

           Arrays.sort(people, new Person());


           Person[] rst = new Person[n];
           for(Person p : people){
               int count = 0;
               for(int i = 0; i < n ; i++){
                     if(count == p.infront){
                        while(rst[i] != null && i < n - 1){
                            i++;
                        }
                        rst[i] = p;
                        break;
                    }
                    if(rst[i] == null) count++;
               }

            }
            ArrayList heightrst = new ArrayList();
            for(int i = 0; i < n; i++){
                heightrst.add(rst[i].height);
            }
            return heightrst;
        }
    }