1 class Solution { 2 public: 3     string multiply(string num1, string num2) { 4         int m = num1.size(); int n = num2.size(); 5         int *d = new int[m+n]; 6         fill_n(d,m+n,0);     7         reverse(num1.begin(),num1.end()); 8         reverse(num2.begin(),num2.end()); 9         for (int i = 0; i < m; i++)10         {11             for (int j = 0; j < n; j++)12             {13                 d[i + j] +=(num1[i]-‘0‘)*(num2[j]-‘0‘);14             }15         }16         string res("");17         for (int i = 0; i < m + n; i++)18         {19             int digit = d[i] % 10;20             int carry = d[i] / 10;21             res.insert(0,1,char(digit+‘0‘));//注意在插入char元素时的用法。22             if (i < m + n - 1) d[i + 1] += carry;23         }24         //此时有可能前面的（下标小的）一些元素是0，要去除。25         while (res[0] == ‘0‘ && res.length()>1)//是>1,不能是>0，否则结果为0时就删空了。26         {27             res.erase(res.begin());28         }29         return res;30     }31 };