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：

两件事情：1.字符串表示的数字和单个字符表示的数字的乘法 2. 两个字符串表示的数字的求和。

这些都是初中计算机竞赛水平的问题。

需要注意的是积为0的情况，需要去除多余的0，比较麻烦。这里采用一些比较恶心的技巧处理掉了。

题解：

class Solution 
{
public:
    void accumulate_string(string& s1, const string& s2)
    {
        string sum;
        sum.resize(max(s1.size(), s2.size()));
        int riter = sum.size() - 1;
        
        int s1_adjust = sum.size() - s1.size();
        int s2_adjust = sum.size() - s2.size();
    
        char incr = 0;
    
        while(riter >= 0)
        {
            int s1iter = riter - s1_adjust;
            char v1 = (s1iter >= 0 ? s1[s1iter] : '0');
            int s2iter = riter - s2_adjust;
            char v2 = (s2iter >= 0 ? s2[s2iter] : '0');
    
            char s = v1 + v2 - 2 * '0' + incr;
    
            if (s >= 10)
                incr = 1, s -= 10;
            else
                incr = 0;
    
            sum[riter] = s + '0';
            --riter;
        }
    
        s1 = (incr == 0 ? string("") : string("1")) + sum;
    }
    
    void multiple_strch(string& s1, char ch)
    {
        string product;
        product.resize(s1.size());
        
        if (ch == '0')  // avoid "0000" problem
        {
            s1 = "0";
            return;
        }
        
        if (ch == '1') // accelerate
            return;
    
        char chval = ch - '0';
        char incr = 0;
        auto piter = product.rbegin();
        for(auto iter = s1.rbegin(); iter != s1.rend(); ++iter, ++piter)
        {
            char val = (*iter - '0') * chval + incr;
            incr = val / 10;
            val -= incr * 10;
            val += '0';
            *piter = val;
        }
        s1 = (incr == 0 ? string("") : string(1, incr + '0')) + product;
    }
    
    string multiply(string& s1, string& s2)
    {
        if (s1.size() < s2.size())
            swap(s1, s2);
            
        string sum="0";
        int i = 0;
        for(auto iter = s2.rbegin(); iter != s2.rend(); ++iter, ++i)
        {
            string prod = s1 + string(i, '0');
            multiple_strch(prod, *iter);
            accumulate_string(sum, prod);
        }
        return sum;
    }
};