[leetcode] Find Anagram Mappings

最新推荐文章于 2021-08-20 05:59:15 发布

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最新推荐文章于 2021-08-20 05:59:15 发布

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分类专栏： algorithms

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本文介绍了一种算法，用于解决两个等长数组A和B的问题，其中B是A的字母异位词。该算法的目标是找到从A到B的索引映射P，即确定A中的每个元素在B中对应的位置。

[problem]

Find Anagram Mappings My Submissions

Difficulty: Easy

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 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].

[thinking]

anagram,由颠倒字母顺序而构成的字. 返回颠倒之后所对应的每个字符的顺序。

[solution]

#include <algorithm> 
class Solution {
public:
    vector<int> anagramMappings(vector<int>& A, vector<int>& B) {
        int length=A.size();
        vector<int> C;
         if(length==0 || length==1) 
         {
             C.push_back(0);
             return C;
         }
        std::vector<int>::iterator it;
        for(int i=0;i<length;i++)
        {
            it=find(B.begin(),B.end(),A[i]);
             int j=it-B.begin();
            C.push_back(j);
        }
        return C;
    }
};