Description

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

分析

题目的意思是：给你queries，代表着排列中的数，返回其对应的数字，在每次query后，需要把当前位置的数字放到最开始的位置，然后其他位置的数依次移动到后面。
所以我的思路也很直接，先把这个排列表示出来，用P表示，按照query的顺序每次排列P，即把找到的数提前，其他的移动到后面去。

代码

class Solution:
    def processQueries(self, queries: List[int], m: int) -> List[int]:
        P=[i+1 for i in range(m)]
        res=[]
        for query in queries:
            idx=P.index(query)
            res.append(idx)
            pos=idx
            while(pos>0):
                P[pos]=P[pos-1]
                pos-=1
            P[0]=query
        return res