Find-Anagram-Mappings

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LeetCode#760 Find Anagram Mappings

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[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].

解题思路

  起初我以为是找到第一个数组里的数字在第二个数组里的索引，考虑到These lists A and B may contain duplicates. 这句话，我还是用了一个数组保存这个值是否已经取过来保证结果集里不出现重复的索引。然而我想多了，这道题仅仅是让你二重循环一下。看代码吧

解题代码【.CPP】

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class Solution {
public:
    vector<int> anagramMappings(vector<int>& A, vector<int>& B) {
        vector<int> res(A.size());
        for (int i = 0; i < A.size(); ++i) {
            for (int j = 0; j < B.size(); ++j) {
                if(B[j] == A[i]){
                    res[i] = j;
                }
            }
        }
        return res;
    }
};
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