Description:
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]]

题意：给定一个二维数组，数组的每一行中的第一个元素表示一个人的身高，第二个元素表示在这个数组中身高大于或等于其他人的人数；现在要求我们对这个二维数组进行重新排序，使得满足数组中第二个元素k的要求，即排在此元素之前的人中仅有k个人的身高大于或等于他；

解法一：我们首先按照身高进行排序后，遍历数组，根据数组中的第二个元素进行插入排序(如果有相等的，应当插入到最前方，因为越靠后的元素，其身高越低)；

Java
class Solution {
    public int[][] reconstructQueue(int[][] people) {
        int[][] result = new int[people.length][2];
        if (people == null || people.length == 0 || 
            people[0].length == 0) {
            return result;
        }
        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];
            }
        });
        
        ArrayList<int[]> temp = new ArrayList<>();
        for (int[] p: people) {
            temp.add(p[1], p);
        }
        int i = 0;
        for (int[] p: temp) {
            result[i] = p;
            i++;
        }
        
        return result;
    }
}

解法二：与解法一的方法是相同的，只是在排序中使用到了Lambda表达式，及ArrayList类中的一个toArray()方法；

Java
class Solution {
    public int[][] reconstructQueue(int[][] people) {
        Arrays.sort(people, (a, b) -> (a[0] == b[0] ? 
                    a[1] - b[1] : b[0] - a[0]));
        ArrayList<int[]> temp = new ArrayList<>();
        for (int[] p: people) {
            temp.add(p[1], p);
        }
        return temp.toArray(new int[0][0]);
    }
}