Description
Given two lists Aand B, and B is an anagram of A. B is an anagram of A means B is made by randomizing the order of the elements in A.
We want to find an index mapping P, from A to B. A mapping P[i] = j means the ith element in A appears in B at index j.
These lists A and B may contain duplicates. If there are multiple answers, output any of them.
For example, given
A = [12, 28, 46, 32, 50] B = [50, 12, 32, 46, 28]We should return
[1, 4, 3, 2, 0]
as P[0] = 1 because the 0th element of A appears at B[1], and P[1] = 4 because the 1st element of A appears at B[4], and so on.
Solution
def anagramMappings(self, A, B):
"""
:type A: List[int]
:type B: List[int]
:rtype: List[int]
"""
C={value:index for index,value in enumerate(B)}
D=[C[x] for x in A]
return D
本文介绍了一种解决两个列表A和B之间的索引映射问题的方法,其中B是A的一个乱序版本。通过构建一个从B到其索引的映射表,可以有效地找出A中每个元素在B中的对应位置。
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