LeetCode 760. Find Anagram Mappings

原题链接在这里：https://leetcode.com/problems/find-anagram-mappings/description/

题目：

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].

题解：

用map保存B的元素与对应位置. 再用A的元素在map中找对应位置.

Time Complexity: O(A.length).

Space: O(1), regardless res.

AC Java:

 class Solution {
     public int[] anagramMappings(int[] A, int[] B) {
         if(A == null || B == null || A.length != B.length){
             throw new IllegalArgumentException("Invalid input.");
         }

         HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();
         for(int i = 0; i<B.length; i++){
             hm.put(B[i], i);
         }

         int [] res = new int[A.length];
         for(int i = 0; i<A.length; i++){
             res[i] = hm.get(A[i]);
         }

         return res;
     }
 }