A,B中元素相同，数组A的下标映射到数组B，Find Anagram Mappings

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原文链接：https://my.oschina.net/liyurong/blog/1616712

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本文介绍了一种通过哈希映射解决数组元素匹配的问题。给定两个数组A和B，B是A的乱序版本，文章提供了一个Java实现的例子，通过创建哈希表来存储数组B中每个元素的位置，然后遍历数组A，找到每个元素在B中的位置，最终返回一个映射列表。

2019独角兽企业重金招聘Python工程师标准>>>

问题：

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:

A, B have equal lengths in range [1, 100].
A[i], B[i] are integers in range [0, 10^5].

解决：

【题意】给两个数组，数组A和数组B中有相同的数字，把数组A中每个数字在数组B中对应位置的下标输出。

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;
}
}