760. Find Anagram Mappings

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 Aappears 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.

解题要点：

以HashMap的方式来存储B的数字和index，然后遍历A，跟map里对应找出B的index，存到一个新的数组里，并返回它。

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];
        int j = 0;
        for(int a : A){
            res[j++] = map.get(a);
        }
        return res;
    }
}