描述
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]
Note:
- 1 <= m <= 10^3
- 1 <= queries.length <= m
- 1 <= queries[i] <= m
解析
根据题意，先构造数组 p ，然后遍历 queries ，从 p 中找出 queries[i] 的索引进行保存到 r ，然后将 queries[i] 移到 p 的开头，重复这个操作直到遍历 queries 结束，返回 r 。
解答
class Solution(object):
def processQueries(self, queries, m):
"""
:type queries: List[int]
:type m: int
:rtype: List[int]
"""
p = [i+1 for i in range(m)]
r = []
for i in queries:
r.append(p.index(i))
p.remove(i)
p = [i] + p
return r
运行结果
Runtime: 52 ms, faster than 49.37% of Python online submissions for Queries on a Permutation With Key.
Memory Usage: 13.7 MB, less than 42.41% of Python online submissions for Queries on a Permutation With Key.
原题链接：leetcode.com/problems/qu…
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