leetcode406 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]]

1、先把Input按照从高到低排序，如果身高相同，按照K值升序。

2、排序后的数组中的每一个一维数组a中的a[1]作为result队列中的index。

[[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]

第一步排序后

[[7,0], [7,1], [6,1], [5,0], [5,2], [4,4]]

向result中插入：

[[7,0]] //0为index

[[7,0],[7,1]] //1为index

[[7,0],[6,1],[7,1]] //1为index

.....

Java代码：

 public static int[][] resconstructQueue(int[][] people) {
  if (people == null) return null;
  Arrays.sort(people,new Comparator<int[]>(){//按照身高降序排序 如果身高相同，按照K值升序排序。
   @Override
   public int compare(int[] o1, int[] o2) {
    if(o1[0] > o2[0]) return -1;
    if(o1[0] < o2[0]) return 1;
    else{
     if(o1[1] > o2[1]) return 1;
     if(o1[1] < o2[1]) return -1;
     else{
      return 1;
     }
    }
   }
  });
  for(int i=0; i<people.length; i++) {
   for(int j=0; j<2;j++) {
    System.out.println(people[i][j]);
   }
  }
  List<int[]> list = new ArrayList<>();
  for(int[] a : people) {
   list.add(a[1],a);
  }
  int[][] res = new int[people.length][2];
  for(int i=0; i<list.size(); i++) {
   res[i] = list.get(i);
  }
  return res;
 }

上边代码排序那块可以优化一下：

        Arrays.sort(people, new Comparator<int[]>() {
            public int compare(int[] a, int[] b) {
                if (b[0] == a[0]) return a[1] - b[1];
                return b[0] - a[0];
            }
        });