python中enumerate()函式運用
昨天在leetcode上做題，發現答案的解法特別簡潔，特此記錄下來，來感受一下python之美～
題目要求如下：
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
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.
Note:
A, Bhave equal lengths in range[1, 100].A[i], B[i]are integers in range[0, 10^5].
class Solution(object):
def anagramMappings(self, A, B):
D = {x: i for i, x in enumerate(B)}
return [D[x] for x in A]
做題時還在苦逼的寫著for迴圈，看到答案的一刻發現自己真是有點弱，anyway，繼續加油！
解釋一下這個優美的地方，首先介紹一下enumerate(),他是python的內建函式，可以對一個可迭代的（iterable）或者可遍歷的物件（如列表、字串），將其組成一個索引序列，利用它可以同時獲得索引和值，解法中將B變成了一個key：value的字典，然後把value和key對換組成一個新的字典D，return [D[x] for x in A]的意思是，如果新字典的value值在A中就返回其key值，就得到了A與B中元素的對映關係，真的很簡潔。