LeetCode 760 - Find Anagram Mappings

https://www.cnblogs.com/grandyang/p/8570939.html

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.

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 Aappears 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]

https://nifannn.github.io/2018/03/06/Algorithm-Notes-Leetcode-760-Find-Anagram-Mappings/
https://segmentfault.com/a/1190000017103654

public int[] anagramMappings(int[] A, int[] B) {
    Map<Integer, Integer> lookup = new HashMap();

    for(int i = 0; i < B.length; i++)
        lookup.put(B[i], i);

    int[] ans = new int[A.length];
    for(int j = 0; j < ans.length; j++)
        ans[j] = lookup.get(A[j]);

    return ans;
}

这道题给了我们两个数组A和B，说是A和B中的数字都相同，但是顺序不同，有点类似错位词的感觉。让我们找出数组A中的每个数字在数组B中的位置。这道题没有太大的难度，用个HashMap建立数组B中的每个数字和其位置之间的映射，然后遍历数组A，在HashMap中查找每个数字的位置即可