LeetCode | 760. Find Anagram Mappings 简单循环题

本文介绍了一种算法，用于寻找两个相同元素集合但顺序不同的数组之间的索引映射关系。通过遍历数组A并查找其元素在数组B中的对应位置，从而建立从A到B的索引映射。

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Giventwo lists Aand B, and B is an anagram of A. B is an anagram of A means B is made by randomizing theorder of the elements in A.

Wewant 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.

Theselists A and B may contain duplicates. Ifthere are multiple answers, output any of them.

Forexample, 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 appearsat B[1], and P[1] = 4 because the 1st element of A appearsat B[4], and so on.

Note:

1. A, B have equal lengths inrange [1, 100].

2. A[i], B[i] are integers in range [0, 10^5].

问你数组A中的每个数在数组b中的索引，遍历一遍A对于每个A，遍历一遍B就好了

class Solution {
public:
    vector<int> anagramMappings(vector<int>& A, vector<int>& B) {
        vector<int> result;
        
        for(int i=0;i<A.size();i++)
        {
            for(int j=0;j<B.size();j++)
            {
                if(B[j]==A[i]) {result.push_back(j);break;}
            }
        }
        return result;
    }
};