43. Multiply Strings

/*
Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2.

Note:

The length of both num1 and num2 is < 110.
Both num1 and num2 contains only digits 0-9.
Both num1 and num2 does not contain any leading zero.
You must not use any built-in BigInteger library or convert the inputs to integer directly.
思路：
乘数与被乘数都从低位开始，一位一位的进行相乘，注意进位
时间复杂度：O(n^2)
*/

#include <stdio.h>
#include <iostream>
#include <string>
#include <sstream>
#include <vector>
#include <algorithm>
#include <string>
#include <cstdlib>
using namespace std;

class Solution {
public:
string multiply(string num1, string num2) {
string sum(num1.size()+num2.size(),'0');
for(int i=num1.size()-1;i>=0;i--)
{
int carr=0;
for(int j=num2.size()-1;j>=0;j--)
{
int tmp=sum[i+j+1]-'0' + (num1[i]-'0')*(num2[j]-'0')+carr;
sum[i+j+1]=tmp%10+'0';
carr=tmp/10;
}
sum[i]+=carr;
}
size_t startPos=sum.find_first_not_of("0");
if(string::npos != startPos)
return sum.substr(startPos);
return "0";
}

};

int main()
{
char c='1';
int t=2;
cout<<c<<endl;
c=t+'0';//一个整数加上一个字符,字符转化为ASIC码与整数相加，相加的结果转化为字符
cout<<c<<endl;
Solution mys;
string s1="11";
string s2="11";
cout<<mys.multiply(s1,s2)<<endl;
return 0;
}