Leetcode-406. Queue Reconstruction by Height 根据身高重建队列 -python

题目

假设有打乱顺序的一群人站成一个队列。 每个人由一个整数对(h, k)表示，其中h是这个人的身高，k是排在这个人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。
链接：https://leetcode.com/problems/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]]

思路及代码

贪心算法

• 从个子高的人开始排，插入到对应的index即可
• 因为个子矮的相对于个子高的人是看不见的，所以也直接插入就行，个子高的就顺位往后挪了

class Solution:
    def reconstructQueue(self, people: List[List[int]]) -> List[List[int]]:
        if not people:
            return []
        peopledict = collections.defaultdict(list)
        for i in range(len(people)):
            peopledict[people[i][0]].append(people[i][1])
        ans = []
        height = sorted(peopledict.keys(),reverse = True)
        for h in height:
            pos = sorted(peopledict[h])
            for position in pos:
                ans.insert(position, [h,position])
        return ans

复杂度

T = $O(n)$
S = $O(n)$