LeetCode | 0406. Queue Reconstruction by Height根据身高重建队列【Python】

LeetCode 0406. Queue Reconstruction by Height根据身高重建队列【Medium】【Python】【贪心】

Problem

LeetCode

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]]

问题

力扣

假设有打乱顺序的一群人站成一个队列。 每一个人由一个整数对 (h, k) 表示，其中 h 是这我的的身高，k 是排在这我的前面且身高大于或等于 h 的人数。 编写一个算法来重建这个队列。

注意：
总人数少于1100人。

示例

输入:
[[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]

输出:
[[5,0], [7,0], [5,2], [6,1], [4,4], [7,1]]

思路

贪心

首先按照身高 h 从高到低，k 从小到大排序。

而后只需插入便可，能够看代码注释。

这里拿示例来举例：

输入：
[[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]

排序：
[[7,0], [7,1], [6,1], [5,0], [5,2], [4,4]]

好比[6,1]，表示前面比 6 高的只有一个，那么天然就插入到位置1（从0开始数）：
[[7,0], [6,1], [7,1], [5,0], [5,2], [4,4]]

以此类推

[5,0]：
[[5,0], [7,0], [6,1], [7,1], [5,2], [4,4]]

[5,2]：
[[5,0], [7,0], [5,2], [6,1], [7,1], [4,4]]

[4,4]：
[[5,0], [7,0], [5,2], [6,1], [4,4], [7,1]]

end

时间复杂度: O(len(people))
空间复杂度: O(1)

Python代码

class Solution(object):
    def reconstructQueue(self, people):
        """
        :type people: List[List[int]]
        :rtype: List[List[int]]
        """
        people.sort(key = lambda x : (-x[0], x[1]))  # 按照h从高到低，k从小到大排序
        res = []
        for p in people:
            res.insert(p[1], p)  # 每次只要在p[1]位置插入p就行，由于p[1]表示p前只能出现的个数
        return res

代码地址

GitHub连接