LeetCode题解之Multiply Strings
1、题目描述

2、问题分析
按照手算乘法的过程进行计算,首先计算乘法,然后计算加法。
3、代码
string multiply(string num1, string num2) {
string small ;
string big;
if( num1.size() < num2.size() ){
small = num1;
big = num2;
}else if( num2.size() < num1.size() ){
small = num2;
big = num1;
}else{
small = num1;
big = num2;
}
int n = small.size() ;
vector<string> re;
string::reverse_iterator it_s = small.rbegin() ;
while( it_s != small.rend() ){
string s;
int up = ;
string::reverse_iterator it_b = big.rbegin() ;
while( it_b != big.rend() ){
int m = (*it_s - '') * (*it_b - '') + up;
if( m < ){
s = std::to_string(m) + s;
up = ;
}else{
s = std::to_string( m% ) + s;
up = m/;
}
++it_b;
}
if( up != ){
s = std::to_string( up ) + s;
}
re.push_back( s );
++it_s;
}
string zero;
for( vector<string>::iterator it = re.begin() ; it != re.end() ; it++ ){
*it += zero;
zero += "";
}
string result = "" ;
for( int i = ; i < re.size() ; i++ ){
result = binaryAdd( result , re[i] );
}
int q = ;
for( string::iterator it = result.begin() ; it != result.end() ; ++it ){
q += (*it - '');
}
if( q == )
result = "";
return result;
}
string binaryAdd( string s1 , string s2 ){
string s;
string::reverse_iterator it1 = s1.rbegin();
string::reverse_iterator it2 = s2.rbegin();
int up = ;
while( it1 != s1.rend() && it2 != s2.rend() ){
int a = (*it1 - '') + (*it2 - '') + up;
if( a < ){
s = std::to_string(a) + s;
up = ;
}else{
s = std::to_string( a - ) + s;
up = ;
}
++it1 ;
++it2 ;
}
while( it1 != s1.rend() ){
if( *it1 - '' + up < ){
s = std::to_string( *it1 - '' + up ) + s;
up = ;
}else{
s = std::to_string( *it1 - '' + up - ) + s;
up = ;
}
++it1;
}
while( it2 != s2.rend() ){
if( *it2 - '' + up < ){
s = std::to_string( *it2 - '' + up ) + s;
up = ;
}else{
s = std::to_string( *it2 - '' + up - ) + s;
up = ;
}
++it2;
}
if( up == ){
s = std::to_string() + s;
}
return s;
}
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