760. Find Anagram Mappings

最新推荐文章于 2019-09-18 17:15:45 发布

原创最新推荐文章于 2019-09-18 17:15:45 发布 · 369 阅读

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本文介绍了一种通过哈希表实现的算法，该算法用于找出两个相同元素但顺序不同的列表之间的映射关系。具体来说，就是将一个列表中的元素与另一个列表中的对应元素建立联系，并返回这种联系的索引映射。

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].

class Node {
    public:
        int  index;
        int  element;
        Node* next;
};
class Solution {
public:
    Node *hashtable[100];
    Solution()
    {         
        for(int i=0;i<100;i++)
        {
            hashtable[i]=NULL;
        }

    }
    vector<int> anagramMappings(vector<int>& A, vector<int>& B) {
        vector<int> i_map;
 

        for(int i=0;i<B.size();i++)
        {
            int hashValue = hash(B[i]);
            Node *e = new Node();
            e->index=i;
            e->element=B[i];
            //头插法
            e->next=hashtable[hashValue];
            hashtable[hashValue]=e;
        }
        for(int i=0;i<A.size();i++)
        {
            int hashValue = hash(A[i]);
            Node *before;
            for(Node *p=hashtable[hashValue]; NULL!=p ; p=p->next)
            {       
                    before=p;
                    if(p->element == A[i])
                    {  i_map.push_back(p->index);
                    //删除找到对应的元素
                    if(hashtable[hashValue] == p)
                    {
                       hashtable[hashValue]=p->next; 
                      
                    }else{
                       before->next=p->next;  
                       
                    }
                    delete p;
                    break;
                    }                           
            }           
            
        }
        return i_map;
        
    }
    int hash( int e){
        return e%100;
    }
    
};