题目地址：https://leetcode.com/problems/find-anagram-mappings/description/open in new window
题目描述：
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.
Wewant 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.
Forexample, 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 B1open in new window, and P1open in new window = 4 because the 1st element of A appears at B[4], and so on. Note:
A,B have equal lengths in range [1, 100]. A[i], B[i] are integers in range [0, 10^5].
Ways
就是找到A中每个元素在B中的位置即可，如果出现了重复的元素，可以返回任意一种次序即可。
方法一：
class Solution:
def anagramMappings(self, A, B):
"""
:type A: List[int]
:type B: List[int]
:rtype: List[int]
"""
answer = []
for a in A:
for i,b in enumerate(B):
if a == b:
answer.append(i)
break
return answer
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方法二：
class Solution:
def anagramMappings(self, A, B):
"""
:type A: List[int]
:type B: List[int]
:rtype: List[int]
"""
return [B.index(a) for a in A]
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方法三：
class Solution:
def anagramMappings(self, A, B):
"""
:type A: List[int]
:type B: List[int]
:rtype: List[int]
"""
d ={}
for i,b in enumerate(B):
d[b] = i
return [d[a] for a in A]
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Date
2018 年 1 月 13 日
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