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:
-
A, Bhave equal lengths in range[1, 100]. -
A[i], B[i]are integers in range[0, 10^5].
Solution
HashMap
class Solution {
public int[] anagramMappings(int[] A, int[] B) {
Map<Integer, List<Integer>> 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;
}
}
