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]]
refer to: https://discuss.leetcode.com/topic/60394/easy-concept-with-python-c-java-solution
- 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]]
1 public class Solution { 2 public int[][] reconstructQueue(int[][] people) { 3 //pick up the tallest guy first 4 //when insert the next tall guy, just need to insert him into kth position 5 //repeat until all people are inserted into list 6 Arrays.sort(people,new Comparator<int[]>(){ 7 @Override 8 public int compare(int[] o1, int[] o2){ 9 return o1[0]!=o2[0]? o2[0]-o1[0] : o1[1]-o2[1]; 10 } 11 }); 12 List<int[]> res = new LinkedList<>(); 13 for(int[] cur : people){ 14 res.add(cur[1],cur); 15 } 16 return res.toArray(new int[people.length][]); 17 } 18 }`