[LeetCode] Find Anagram Mappings 寻找异构映射-优快云博客

本文介绍了一种通过使用HashMap来解决寻找数组A中元素在数组B中的位置的问题的方法。两个数组长度相等且B为A的乱序版本。通过创建一个映射表，将B中的每个元素与它在B中的索引对应起来，从而快速找到A中每个元素在B中的位置。

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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 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 Aappears 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].

这道题给了我们两个数组A和B，说是A和B中的数字都相同，但是顺序不同，有点类似错位词的感觉。让我们找出数组A中的每个数字在数组B中的位置。这道题没有太大的难度，用个HashMap建立数组B中的每个数字和其位置之间的映射，然后遍历数组A，在HashMap中查找每个数字的位置即可，参见代码如下：

class Solution {
public:
    vector<int> anagramMappings(vector<int>& A, vector<int>& B) {
        vector<int> res;
        unordered_map<int, int> m;
        for (int i = 0; i < B.size(); ++i) m[B[i]] = i;
        for (int num : A) res.push_back(m[num]);
        return res;
    }
};