Given two non-negative integers num1
and num2
represented as strings, return the product of num1
and num2
.
Note:
- 1.The length of both
num1
andnum2
is < 110. - 2.Both
num1
andnum2
contains only digits 0-9. - 3.Both
num1
andnum2
does not contain any leading zero.
You must not use any built-in BigInteger library or convert the inputs to integer directly.
典型的大整数乘法模拟
class Solution {
void CharToInt(string &str, vector<short int> &arr)
{
for(int i=str.size()-1; i>=0; -- i)
arr.push_back(str[i]-'0');
}
void sum(vector<short int> &res, vector<short int> &ans)
{
int add = 0;
for(int i=0; i<res.size(); ++ i)
{
ans[i] = res[i] + ans[i] + add;
add = ans[i] / 10;
ans[i] %= 10;
}
int i = res.size();
while(add != 0)
{
ans[i] = ans[i] + add;
add = ans[i] / 10;
ans[i] %= 10;
}
}
public:
string multiply(string num1, string num2) {
vector<short int> a, b;
CharToInt(num1, a);
CharToInt(num2, b);
vector<short int> ans(12500, 0);
for(int i=0; i<a.size(); ++ i)
{
vector<short int> res;
for(int j=0; j<i; ++ j)
res.push_back(0);
int add = 0;
for(int j=0; j<b.size(); ++ j)
{
int t = a[i] * b[j] + add;
res.push_back(t % 10);
add = t / 10;
}
if(add != 0)
res.push_back(add);
sum(res, ans);
}
string str;
bool is = false;
for(int i=12499; i>=0; -- i)
{
if(ans[i] != 0)
is = true;
if(is)
str += ans[i] + '0';
}
if(is == false)
return "0";
return str;
}
};