406 Queue Reconstruction by Height

406. Queue Reconstruction by Height

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]]

分析& 代码

来自discuss

把给的[h, k]对进行排序：

• 如果height一样，按照k的递增排
• 如果height不一样，按照height的递减排

比如，输入[[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]] 排序完应该是这样[7, 0], [7, 1], [6, 1], [5, 0], [5, 2], [4, 4].

整体代码的思路是：

• 先找到最高的所有对，然后按照k的顺序排，比如[7,0], [7,1]
• 然后把次高的对，按照k的顺序放进去，[7,0], [6,1], [7,1]
• 重复

代码是：

public int[][] reconstructQueue(int[][] people) {
    Arrays.sort(people, new Comparator<int[]>() {
        public int compare(int[] a, int[] b) {
            if(a[0] == b[0]) {
                return a[1] - b[1];
            } else {
                return b[0] - a[0];
            }
        }
    });
    List<int[]> list = new ArrayList<>();
    for(int[] person: people) {
        list.add(person[1], person);
    }
    return list.toArray(new int[list.size()][]);
}