760 Find Anagram Mappings

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题目

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

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].

思路

用一个Map把B里面的元素和它的位置存起来，然后便利一遍A就好

代码

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// Java Solution
class Solution {
    public int[] anagramMappings(int[] A, int[] B) {
        Map<Integer,Integer> m = new HashMap<Integer,Integer>();
        for(int i=0;i<B.length;i++) {
            m.put(B[i],i);
        }
        int[] result = new int[A.length];
        for (int i=0;i<A.length;i++) {
            result[i] = m.get(A[i]);
        }
        return result;
    }
}
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//C++ Solution
class Solution {
public:
    vector<int> anagramMappings(vector<int>& A, vector<int>& B) {
        map<int,int> a;
        int counter = 0;
        for(int i:B) {
            a[i] = counter++;
        }
        vector<int> result;
        for(int i:A) {
            result.push_back(a[i]);
        }
        return result;
    }
};