[LeetCode] Find Anagram Mappings 寻找异构映射
2023-09-11 14:21:37 时间
Given two lists A
and 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 i
th 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 0
th element of A
appears at B[1]
, and P[1] = 4
because the 1
st 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]
.
这道题给了我们两个数组A和B,说是A和B中的数字都相同,但是顺序不同,有点类似错位词的感觉。让我们找出数组A中的每个数字在数组B中的位置。这道题没有太大的难度,用个HashMap建立数组B中的每个数字和其位置之间的映射,然后遍历数组A,在HashMap中查找每个数字的位置即可,参见代码如下:
class Solution { public: vector<int> anagramMappings(vector<int>& A, vector<int>& B) { vector<int> res; unordered_map<int, int> m; for (int i = 0; i < B.size(); ++i) m[B[i]] = i; for (int num : A) res.push_back(m[num]); return res; } };
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