LeetCode(43. Multiply Strings)

题目：

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.

思路比较简单，不多说啦

几个需要注意的点：

1. 高位多个为0的，需要进行判断

2. 如果乘数为0的话，可以直接跳过，能节省不少时间

3. string 的两个使用

　1) 创建有 len 长的 ‘0’串  string str(len, '0')

　2) string 的反转 reverse(str.begin(), str.end())  反转后操作好像更容易理解~

 class Solution {
 public:
     string multiply(string num1, string num2) {
         int size1 = num1.size();
         int size2 = num2.size();
         string result(size1 + size2 + , '');
         reverse(num1.begin(),num1.end());
         reverse(num2.begin(), num2.end());
         for(int j = ; j < size2; ++j){//num2
             if(num2[j] == '')
                 continue;
             int k = j;
             for(int i = ; i < size1; ++i){//num1
                 int tmp = (num1[i] - '') * (num2[j] - '') + result[k] - '';
                 result[k] = tmp %  + '';
                 result[k + ] = result[k + ] + tmp / ;
                 ++k;
             }
         }
         while(!result.empty() && result.back() == '') result.pop_back();
         if(result.empty()) return "";
         reverse(result.begin(), result.end());
         return result;
     }
 };