LeetCode 1409. Queries on a Permutation With Key查询带键的排列【Medium】【Python】【模拟】

Problem

LeetCode

Given the array queries of positive integers between 1 and m, you have to process all queries[i] (from i=0 to i=queries.length-1) according to the following rules:

  • In the beginning, you have the permutation P=[1,2,3,...,m].
  • For the current i, find the position of queries[i] in the permutation P (indexing from 0) and then move this at the beginning of the permutation P. Notice that the position of queries[i] in P is the result for queries[i].

Return an array containing the result for the given queries.

Example 1:

Input: queries = [3,1,2,1], m = 5
Output: [2,1,2,1] 
Explanation: The queries are processed as follow: 
For i=0: queries[i]=3, P=[1,2,3,4,5], position of 3 in P is 2, then we move 3 to the beginning of P resulting in P=[3,1,2,4,5]. 
For i=1: queries[i]=1, P=[3,1,2,4,5], position of 1 in P is 1, then we move 1 to the beginning of P resulting in P=[1,3,2,4,5]. 
For i=2: queries[i]=2, P=[1,3,2,4,5], position of 2 in P is 2, then we move 2 to the beginning of P resulting in P=[2,1,3,4,5]. 
For i=3: queries[i]=1, P=[2,1,3,4,5], position of 1 in P is 1, then we move 1 to the beginning of P resulting in P=[1,2,3,4,5]. 
Therefore, the array containing the result is [2,1,2,1].  

Example 2:

Input: queries = [4,1,2,2], m = 4
Output: [3,1,2,0]

Example 3:

Input: queries = [7,5,5,8,3], m = 8
Output: [6,5,0,7,5]

Constraints:

  • 1 <= m <= 10^3
  • 1 <= queries.length <= m
  • 1 <= queries[i] <= m

问题

力扣

给你一个待查数组 queries ,数组中的元素为 1 到 m 之间的正整数。 请你根据以下规则处理所有待查项 queries[i](从 i=0 到 i=queries.length-1):

  • 一开始,排列 P=[1,2,3,...,m]。
  • 对于当前的 i ,请你找出待查项 queries[i] 在排列 P 中的位置(下标从 0 开始),然后将其从原位置移动到排列 P 的起始位置(即下标为 0 处)。注意, queries[i] 在 P 中的位置就是 queries[i] 的查询结果。

请你以数组形式返回待查数组 queries 的查询结果。

示例 1:

输入:queries = [3,1,2,1], m = 5
输出:[2,1,2,1] 
解释:待查数组 queries 处理如下:
对于 i=0: queries[i]=3, P=[1,2,3,4,5], 3 在 P 中的位置是 2,接着我们把 3 移动到 P 的起始位置,得到 P=[3,1,2,4,5] 。
对于 i=1: queries[i]=1, P=[3,1,2,4,5], 1 在 P 中的位置是 1,接着我们把 1 移动到 P 的起始位置,得到 P=[1,3,2,4,5] 。 
对于 i=2: queries[i]=2, P=[1,3,2,4,5], 2 在 P 中的位置是 2,接着我们把 2 移动到 P 的起始位置,得到 P=[2,1,3,4,5] 。
对于 i=3: queries[i]=1, P=[2,1,3,4,5], 1 在 P 中的位置是 1,接着我们把 1 移动到 P 的起始位置,得到 P=[1,2,3,4,5] 。 
因此,返回的结果数组为 [2,1,2,1] 。

示例 2:

输入:queries = [4,1,2,2], m = 4
输出:[3,1,2,0]

示例 3:

输入:queries = [7,5,5,8,3], m = 8
输出:[6,5,0,7,5]

提示:

  • 1 <= m <= 10^3
  • 1 <= queries.length <= m
  • 1 <= queries[i] <= m

思路

模拟

时间复杂度: O(n),n 为 queries 的长度
空间复杂度: O(n)

Python3代码
from typing import List

class Solution:
    def processQueries(self, queries: List[int], m: int) -> List[int]:
        p = [x for x in range(1, m + 1)]
        res = []

        for x in queries:
            temp = p.index(x)
            num = p[temp]
            res.append(temp)
            p.remove(p[temp])
            p.insert(0, num)
        return res

GitHub链接

Python