A,B中元素相同，數組A的下標映射到數組B，Find Anagram Mappings

問題：

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].

解決：

【題意】給兩個數組，數組A和數組B中有相同的數字，把數組A中每一個數字在數組B中對應位置的下標輸出。

①  使用hashmap。

class Solution { //6ms     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 ++){             if (map.containsKey(A[i])){                 int index = map.get(A[i]);                 res[i] = index;             }         }         return res;     } }