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字符串的大数乘法，模拟题

把两个字符串倒过来以后对应位置相乘，注意最终结果要去掉最前面的0

class Solution {
public:
    string multiply(string num1, string num2) {
        reverse(num1.begin(), num1.end());
        reverse(num2.begin(), num2.end());
        string s(num1.size() + num2.size(), ‘0‘);
        for (size_t i = 0; i < num1.size(); i++)
        {
            for (size_t j = 0; j < num2.size(); j++)
            {
                s[i + j] = s[i + j] - ‘0‘ + (num2[j] - ‘0‘)*(num1[i] - ‘0‘);
                s[i + j + 1] = s[i + j + 1] + s[i + j] / 10;
                s[i + j] = s[i + j] % 10 + ‘0‘;
            }
        }
        int notZeroIndex = s.size() - 1;
        while (s[notZeroIndex] == ‘0‘ && notZeroIndex>0)
        {
            notZeroIndex--;
        }
        s = s.substr(0, notZeroIndex + 1);
        reverse(s.begin(), s.end());
        return s;
    }
};

Multiply Strings

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原文地址：http://www.cnblogs.com/flyjameschen/p/4381226.html