结论:下标访问效率更高。

迭代器访问:

用时:超过1600ms

class Solution {
public:
    vector<int> fairCandySwap(vector<int>& A, vector<int>& B) 
    {
       
       int sumA=0;
       for(int i=0;i<A.size();i++)
       {
           sumA+=A[i];
       }
    
       
       int sumB=0;
       for(int i=0;i<B.size();i++)
       {
           sumB+=B[i];
       }

        int diff=sumA-sumB;
        //cout<<"diff="<<diff<<endl;
        vector<int> ret;
        bool isSwap=false;
        for(int a=0;a<A.size();a++)
        {
            for(int b=0;b<B.size();b++)
            {//cout<<"a-b="<<*itera-*iterb<<endl;
                if(diff==2*(A[a]-B[b]))
                {                    
                    ret.push_back(A[a]);
                    ret.push_back(B[b]);
                    isSwap=true;
                    break;
                }
            }
            if(isSwap==true)
                break;


        }  
        return ret;
    }   
};

 

下标访问:
用时:952ms
class Solution {
public:
    vector<int> fairCandySwap(vector<int>& A, vector<int>& B) 
    {
       
       int sumA=0;
       for(int i=0;i<A.size();i++)
       {
           sumA+=A[i];
       }
    
       
       int sumB=0;
       for(int i=0;i<B.size();i++)
       {
           sumB+=B[i];
       }

        int diff=sumA-sumB;
        //cout<<"diff="<<diff<<endl;
        vector<int> ret;
        bool isSwap=false;
        for(int a=0;a<A.size();a++)
        {
            for(int b=0;b<B.size();b++)
            {//cout<<"a-b="<<*itera-*iterb<<endl;
                if(diff==2*(A[a]-B[b]))
                {                    
                    ret.push_back(A[a]);
                    ret.push_back(B[b]);
                    isSwap=true;
                    break;
                }
            }
            if(isSwap==true)
                break;

        }  
        return ret;
    }   
};

 

 

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