406. Queue Reconstruction by Height

最新推荐文章于 2023-11-18 01:25:37 发布

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最新推荐文章于 2023-11-18 01:25:37 发布

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分类专栏： LeetCode Greedy 文章标签： LeetCode Greedy Queue Reconstruction

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本文介绍了一种用于根据人员的高度及前面高于或等于其高度的人数来重构队列的算法。通过将人群分为不同高度组并逐步插入的方式，最终实现队列的正确排序。

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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]]

非常好理解的思路：

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]]

代码如下：

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

按下面的方法，重写compare函数速度会比较快。代码如下：

public class Solution {
    public int[][] reconstructQueue(int[][] people) {
        Arrays.sort(people,new Comparator<int[]>(){
           @Override
           public int compare(int[] o1, int[] o2){
               return o1[0]!=o2[0]?-o1[0]+o2[0]:o1[1]-o2[1];
           }
        });
        List<int[]> res = new LinkedList<>();
        for(int[] cur : people){
            res.add(cur[1],cur);       
        }
        return res.toArray(new int[people.length][]);
    }
}