【Multiply Strings】cpp

题目：

Given two numbers represented as strings, return multiplication of the numbers as a string.

Note: The numbers can be arbitrarily large and are non-negative.

代码：

class Solution {
public:
    string multiply(string num1, string num2) {
            const int n1 = num1.size(), n2 = num2.size();
            if ( num1=="" || num2=="") return "";
            vector<char> ret;
            for ( int i=n1-; i>=; --i )
            {
                vector<char> local(,'');
                int v = , carry = , curr = ;
                for ( int j=n2-; j>=; --j )
                {
                    v= (num1[i]-'')*(num2[j]-'');
                    carry = v/;
                    curr = v%;
                    carry += ((local[]-'')+curr)/;
                    curr = ((local[]-'')+curr)%;
                    local[] = curr+'';
                    local.insert(local.begin(), carry+'');
                }
                if (local[]=='') local.erase(local.begin());
                // cout << string(local.begin(),local.end()) << endl;
                // add zeros
                for ( int z=n1-; z>i; --z) local.push_back('');
                // cout << string(local.begin(),local.end()) << endl;
                carry = , curr = ;
                for ( int r=ret.size()-, l=local.size()-; r>= || l>=; --r,--l )
                {
                    if ( r>= && l>= )
                    {
                        curr = (carry+(ret[r]-'')+(local[l]-''))%;
                        carry = (carry+(ret[r]-'')+(local[l]-''))/;
                        local[l] = curr+'';
                    }
                    else
                    {
                        curr = (carry+(local[l]-''))%;
                        carry = (carry+(local[l]-''))/;
                        local[l] = curr+'';
                    }
                }
                if (carry!=) { local.insert(local.begin(), carry+''); }
                ret = local;
            }
            return string(ret.begin(),ret.end());
    }
};

tips：

就是一些字符串操作的细节，考虑进位，0这类的细节。

======================================

第二次过这道题，憋了好久。主要是有个思维误区，认为tmp比ret最多多一位；但是，比如52*4这样的例子，2*4=8 50*4=200就不止一位了。

解决了这个误区，代码AC了。

class Solution {
public:
    string multiply(string num1, string num2) {
            if ( num1=="" || num2=="" ) return "";
            vector<char> ret(num2.size(),'');
            for ( int i=num1.size()-; i>=; --i )
            {
                // mupltiply ret
                vector<int> tmp;
                int val = , carry = ;
                for ( int j=num2.size()-; j>=; --j )
                {
                    val = (num2[j]-'')*(num1[i]-'')+carry;
                    tmp.insert(tmp.begin(), val%);
                    carry = val/;
                }
                if ( carry> ) tmp.insert(tmp.begin(), carry);
                for ( int k=; k<num1.size()--i; ++k ) tmp.push_back();
                // merge tmp & ret
                while ( tmp.size()>ret.size() ) ret.insert(ret.begin(),'');
                val = ;
                carry = ;
                for ( int m=tmp.size()-; m>=; --m )
                {
                    val = (ret[m]-'') + tmp[m] + carry;
                    ret[m] = val%+'';
                    carry = val/;
                }
                if ( carry> ) ret.insert(ret.begin(), carry+'');
            }
            return string(ret.begin(), ret.end());
    }
};