[LeetCode] 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]]

根据身高重建队列。题意是给一个二维数组[h, k]，其中h表示每个人的身高，k表示在这个人之前有几个人的身高大于等于这个人。input的顺序是错的，请你根据每个人的[h, k]的状况对input进行正确排序。

思路是先按照h从大到小排序，如果h相同，则k小的在前。排序过后是类似这样。

此时创建一个list，将排序后的结果加入list，加入的规则是将每个人的[h, k]加入到list的k位置上。

时间O(nlogn)

空间O(n^2) - output

Java实现

class Solution {
    public int[][] reconstructQueue(int[][] people) {
        Arrays.sort(people, (a, b) -> a[0] == b[0] ? a[1] - b[1] : b[0] - a[0]);
        // 新建一个List
        List<int[]> list = new ArrayList<>();
        // 将每个数组中第二个数值作为下标，将该数组插入List
        for (int[] p : people) {
            list.add(p[1], p);
        }
        // 将list转为数组返回
        int[][] res = new int[people.length][2];
        for (int i = 0; i < list.size(); i++) {
            res[i] = list.get(i);
        }
        return res;
    }
}

JavaScript实现

/**
 * @param {number[][]} people
 * @return {number[][]}
 */
var reconstructQueue = function (people) {
    let res = [];
    people.sort((a, b) => (a[0] == b[0] ? a[1] - b[1] : b[0] - a[0]));
    people.forEach((val) => {
        res.splice(val[1], 0, val);
    });
    return res;
};

LeetCode 题目总结