760. Find Anagram Mappings - Easy

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].

 

扫一遍B，把<B[i], i>存进hashset，再扫一遍A，拿到对应B中的index

time: O(n), space: O(n)

class Solution {
    public int[] anagramMappings(int[] A, int[] B) {
        HashMap<Integer, Integer> map = new HashMap<>();
        for(int i = 0; i < B.length; i++) {
            map.put(B[i], i);
        }
        int[] res = new int[A.length];
        for(int i = 0; i < A.length; i++) {
            res[i] = map.get(A[i]);
        }
        return res;
    }
}