760.Find Anagram Mappings

#### QUESTION:

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 `i`th 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 = 1
``````

because the

``````0
``````

th element of

``````A
``````

appears at

``````B
``````

, and

``````P = 4
``````

because the

``````1
``````

st element of

``````A
``````

appears at

``````B
``````

, 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]`.

#### EXPLANATION:

1.将b中元素和对应的位置一一取出。因为有重复的，所以用stack进行存储。

2.遍历A中的元素，获取到对应的位置，进行pop，这样就不会选取到两次一样的位置了。

#### SOLUTION:

``````import java.util.*;
class Solution {
public int[] anagramMappings(int[] A, int[] B) {
Hashtable<Integer,Stack<Integer>> table = new Hashtable<>();
for(int i = 0;i<B.length;i++){
Stack<Integer> tmp = table.get(B[i]);
if(tmp==null)
tmp = new Stack<Integer>();
tmp.push(i);
table.put(B[i],tmp);
}
int[] result = new int[A.length];
for(int i = 0;i<A.length;i++){
result[i] = table.get(A[i]).pop();
}
return result;
}
}
``````
>