学步园

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.

思路：模拟大数相乘，比较简单。

class Solution {
public:
    string multiply(string num1, string num2) {
        string s;
        string sum,tmp;
        if(num1.length()<=num2.length())
        {
           s = num1;
           num1 = num2;
           num2 = s;
        }
        int len1 = num1.length();
        int len2 = num2.length();
        int i,j,k,len;
        int flag;
        char ch;
        sum = "0";
        for(i=len2-1; i>=0; --i)
        {
           flag = 0;
           s = "";
           for(k=0; k<len2-1-i; ++k)
             s.push_back('0');
           for(j=len1-1; j>=0; --j)
           {
              ch = '0' + ((num2[i]-'0')*(num1[j]-'0')+flag)%10;
              flag = ((num2[i]-'0')*(num1[j]-'0')+flag)/10;
              s.push_back(ch);
           }
           if (flag > 0)
              s.push_back(flag + '0');
           len = (s.length() >= sum.length() ? s.length() : sum.length());
           flag = 0;
           tmp = "";
           for(k=0; k<len; k++)
           {
                if (k >= s.length())
                {
                      tmp.push_back((sum[k]-'0'+flag)%10 + '0');
                      flag = (sum[k]-'0'+flag)/10;
                }   
                else if (k >= sum.length())
                {
                     tmp.push_back((s[k]-'0'+flag)%10 + '0');
                     flag = (s[k]-'0'+flag)/10;
                } 
                else 
                {
                     tmp.push_back((sum[k]-'0'+flag + s[k]-'0')%10 + '0');
                     flag = (sum[k]-'0'+flag + s[k]-'0')/10;
                }
           }
           if (flag == 1)
              tmp.push_back('1');
           sum = tmp;
        }
        string result(sum.rbegin(), sum.rend());
        s = "";
        len = result.length();
        for(i=0; i<len; i++)
        {
           if (result[i] == '0')
           {
              continue;
           }

          for(j=i; j<len; ++j)
             s.push_back(result[j]);
          break;
        }
        if (s.length() == 0)
           s.push_back('0');
        return s;
    }
};