[LeetCode] 大数问题，相加和相乘，题 Multiply Strings

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.

class Solution {
public:
    string multiply(string num1, string num2) {
    }
};

我的解法是每次提取num2的一位，然后和num1相乘，所得结果并入string res中，这样虽然每次都要新定义一个string，效率略低，但是思路比较清晰：把字符串相乘划分成了“字符串和数字相乘” 和 “字符串相加” 两个子问题，分别解决就可以。

处理时先将num1，和num2逆序，使得最低位在[0] 位置，这样处理起来比较方便，不易写错。

class Solution {
public:
    string multiply(string num1, string num2) {
        if(num1.length() ==  || num2.length() == )
            return "";
        if(num1.length() ==  && num1[] == '') return "";
        if(num2.length() ==  && num2[] == '') return "";
        string res(num1.length() + num2.length(), '');
        int i = , j = ;
        reverse(num1.begin(), num1.end());
        reverse(num2.begin(), num2.end());
        for(i = ; i < num2.length(); ++i){
            int offset = i;
            string tmp(offset + num1.length() + , '');
            int add = ;
            for(j = ; j < num1.length(); ++j){
                int tmpMul = (num1[j] - '') * (num2[i] - '') + add;
                add = tmpMul / ;
                tmp[j + offset] = (tmpMul % ) + '';
            }
            tmp[j + offset] = add + '';
            plus(res, tmp);
        }
        int countStartZero = ; //高位'0'的个数
        for(i = res.length() - ; i >=  && res[i] == ''; --i, ++countStartZero);
        if(countStartZero == res.length()) return "";
        res = res.substr(, res.length() - countStartZero);
        reverse(res.begin(), res.end());
        return res;
    }
private:
    void plus(string &s1, string &s2){
        int add = ;
        for(int i = ; i < s1.length() && (add >  || i < s2.length()); ++i){
            int tmp = (s1[i] - '') + add;
            if(i < s2.length()) tmp += (s2[i] - '');
            add = tmp / ;
            s1[i] = (tmp % ) + '';
        }
    }
};