760. Find Anagram Mappings

Leetcode Hash Table

Given two lists A and 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.

Example

Given A = [12, 28, 46, 32, 50] and B = [50, 12, 32, 46, 28], return [1, 4, 3, 2, 0].

Explanation:
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, B have equal lengths in range [1, 100].
• A[i], B[i] are integers in range [0, 10^5].

分析

这道题目LeetCode收费，没有会员的请前往LintCode。一看到这道题目就会想起LeetCode 205. Isomorphic Strings。在Q205中，确认字符串是否同构的关键步骤是构造一个映射，而这里题目明确指明需要求出这个映射。唯一的区别是，Q205里是字符串，这里是数字。

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

    for (int i = 0; i < A.length; i++)
        res[i] = map.get(A[i]);
    return res;
}