[Leetcode] 760. Find Anagram Mappings 解题报告

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本文介绍了一种算法，用于解决两个相同元素组成的列表之间的索引映射问题。通过使用哈希表记录元素及其在目标列表中的位置，快速找到每个元素对应的映射关系。

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]

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].

思路：

练手题目，哈哈哈。

代码：

class Solution {
public:
    vector<int> anagramMappings(vector<int>& A, vector<int>& B) {
        unordered_map<int, int> hash;   // map from value to index
        for (int i = 0; i < B.size(); ++i) {
            hash[B[i]] = i;
        }
        vector<int> ans(A.size(), 0);
        for (int i = 0; i < A.size(); ++i) {
            ans[i] = hash[A[i]];
        }
        return ans;
    }
};