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 + 1, '0');
        reverse(num1.begin(),num1.end());
        reverse(num2.begin(), num2.end());
        for(int j = 0; j < size2; ++j){//num2
            if(num2[j] == '0')
                continue;
            int k = j;
            for(int i = 0; i < size1; ++i){//num1
                int tmp = (num1[i] - '0') * (num2[j] - '0') + result[k] - '0';
                result[k] = tmp % 10 + '0';
                result[k + 1] = result[k + 1] + tmp / 10;
                ++k;
            }
        }
        while(!result.empty() && result.back() == '0') result.pop_back();
        if(result.empty()) return "0";
        reverse(result.begin(), result.end());
        return result;
    }
};