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 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 Aappears 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].
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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