760. Find Anagram Mappings

Description

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].

Solution

HashMap

class Solution {
    public int[] anagramMappings(int[] A, int[] B) {
        Map> valToIndexes = new HashMap<>();
        for (int i = 0; i < B.length; ++i) {
            if (!valToIndexes.containsKey(B[i])) {
                valToIndexes.put(B[i], new LinkedList<>());
            }
            valToIndexes.get(B[i]).add(i);
        }
        
        int[] mappings = new int[A.length];
        for (int i = 0; i < mappings.length; ++i) {
            mappings[i] = valToIndexes.get(A[i]).get(0);
            valToIndexes.get(A[i]).remove(0);
        }
        
        return mappings;
    }
}