leetcode406. 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]]

解法

对二维数组people进行排序，排序规则为：先按照身高降序排序，如果身高相同则按照第二个数（前面有几个人的个数）升序进行排序，然后再按照第二个数的位置进行插入list列表中，完成之后再复制到新的数组中。

// @author: taotao @date: 17/3/30
public class Solution {
    public int[][] reconstructQueue(int[][] people) {
        if (people.length == 0 || people == null) {
            return people;
        }
        if (people[0].length == 0 || people[0] == null) {
            return people;
        }

        int[][] ret = new int[people.length][2];
        Arrays.sort(people, new Comparator<int[]>() {
            @Override
            public int compare(int[] o1, int[] o2) {
                if (o1[0] == o2[0]) {
                    return o1[1] - o2[1];
                } else {
                    return o2[0] - o1[0];
                }
            }
        });
        List<int[]> temp = new ArrayList<>();
        for (int i = 0; i < people.length; i++) {
            temp.add(people[i][1], people[i]);
        }
        for (int i = 0; i < people.length; i++) {
            ret[i] = temp.get(i);
        }

        return ret;
    }
}