760.Find Anagram Mappings

QUESTION:

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

0

th element of

A

appears at

B[1]

, and

P[1] = 4

because the

1

st 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].

EXPLANATION:

思路也是很容易的

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;
    }
}
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