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

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A = [12, 28, 46, 32, 50]
B = [50, 12, 32, 46, 28]

We should return

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[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:

  1. A, B have equal lengths in range [1, 100].
  2. A[i], B[i] are integers in range [0, 10^5].

遍历一次数组B，记录每个值出现的下标，再遍历A，依次对应起来。

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class Solution {
public:
    vector<int> anagramMappings(vector<int>& A, vector<int>& B) {
        unordered_map<int, queue<int>> m;
        for (int i = 0; i < B.size(); i++) {
            m[B[i]].push(i);
        }

        vector<int> ans;
        for (int i = 0; i < A.size(); i++) {
            ans.push_back(m[A[i]].front());
            m[A[i]].pop();
        }

        return ans;
    }
};