43. Multiply Strings

题目

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2.

Note:

    The length of both num1 and num2 is < 110.
    Both num1 and num2 contains only digits 0-9.
    Both num1 and num2 does not contain any leading zero.
    You must not use any built-in BigInteger library or convert the inputs to integer directly.

解析

• 大数相乘原则
• 认清楚乘法从后面开始乘，vec[size-1]是最小位

string multiply(string num1, string num2) {
    int l1 = num1.size(), l2 = num2.size();
    string res(l1 + l2, '0');
    if (l1 == 0 || l2 == 0)
        return "";
    for (int i = l1 - 1; i >= 0; i--){
        int add = 0;
        for (int j = l2 - 1; j >= 0; j--){
            int mul = (num1[i] - '0')*(num2[j] - '0');
            int sum = res[i + j + 1] + add + mul % 10 - '0';
            res[i + j + 1] = '0' + sum % 10;
            add = mul / 10 + sum / 10;
        }
        res[i] += add;
    }
    for (int i = 0; i < l1 + l2; i++)
    if (res[i] != '0')
        return res.substr(i);
    return "0";
}

int main()
{
    string str1, str2;
    cin >> str1;
    cin >> str2;

    string ret = multiply(str1, str2);

    cout << ret << endl;
    return 0;
}

• 自己实现

// add 43. Multiply Strings
class Solution_43 {
public:
    string multiply(string num1, string num2) {
        int len1 = num1.size();
        int len2 = num2.size();

        vector<int> vec(len1+len2,0); // 初始化内存空间
        //vec.reserve(len1 + len2);
        
        for (int i = 0; i < len1; i++)
        {
            int k = i;
            for (int j = 0; j < len2;j++)
            {
                //vec.push_back(a*b);
                vec[k] += (num1[len1-1-i] - '0')*(num2[len2-1-j]-'0');  ////Calculate from rightmost to left
                k++; 
            }
        }

        string ret="";
        for (int i = 0; i < vec.size();i++)
        {
            if (vec[i]>=10)
            {
                vec[i+1] += vec[i] / 10;
                vec[i] = vec[i] % 10;
            }
            //char temp[5];
            //_itoa(vec[i], temp, 10);
            //ret += temp;

            char temp = vec[i] + '0'; 
            ret += temp; //反着取得
        }

        //reserve(ret.begin(),ret.end());
        //reverse(vec.begin(), vec.end()); //string没有反转函数，vector有
        
        //判断第一个非0位 //size_t startpos = sum.find_first_not_of("0");
        int flag = 0;
        for (int i = ret.size()-1; i >=0;i--)
        {
            if (ret[i]!='0')
            {
                flag = i;
                break;
            }
        }

        int begin = 0, end = flag;
        while (begin < end)
        {
            swap(ret[begin], ret[end]);
            begin++;
            end--;
        }

        return ret.substr(0,flag+1);
    }
};

题目来源

• 43. Multiply Strings