将大数相乘表示为字符串

给定两个正数作为字符串。这些数字可能非常大(可能不适合int long int int)，任务是找到这两个数字的乘积。
例子：

Input : num1 = 4154  
        num2 = 51454
Output : 213739916 

Input :  num1 = 654154154151454545415415454  
         num2 = 63516561563156316545145146514654 
Output : 41549622603955309777243716069997997007620439937711509062916

这个想法是基于学校数学的。

我们从第二个数字的最后一位开始乘以第一个数字。然后，我们将第二个数字的第二个数字与第一个数字相乘，依此类推。我们将所有这些乘法相加。在相加时，我们将第i个乘法移位了。
以下解决方案中使用的方法是仅保留一个数组以获取结果。我们循环遍历所有数字的第一个和第二个数字，并将结果添加到适当的位置。

C++

// C++ program to multiply two numbers represented
// as strings.
#include
using namespace std;
 
// Multiplies str1 and str2, and prints result.
string multiply(string num1, string num2)
{
    int len1 = num1.size();
    int len2 = num2.size();
    if (len1 == 0 || len2 == 0)
    return "0";
 
    // will keep the result number in vector
    // in reverse order
    vector result(len1 + len2, 0);
 
    // Below two indexes are used to find positions
    // in result.
    int i_n1 = 0;
    int i_n2 = 0;
     
    // Go from right to left in num1
    for (int i=len1-1; i>=0; i--)
    {
        int carry = 0;
        int n1 = num1[i] - '0';
 
        // To shift position to left after every
        // multiplication of a digit in num2
        i_n2 = 0;
         
        // Go from right to left in num2            
        for (int j=len2-1; j>=0; j--)
        {
            // Take current digit of second number
            int n2 = num2[j] - '0';
 
            // Multiply with current digit of first number
            // and add result to previously stored result
            // at current position.
            int sum = n1*n2 + result[i_n1 + i_n2] + carry;
 
            // Carry for next iteration
            carry = sum/10;
 
            // Store result
            result[i_n1 + i_n2] = sum % 10;
 
            i_n2++;
        }
 
        // store carry in next cell
        if (carry > 0)
            result[i_n1 + i_n2] += carry;
 
        // To shift position to left after every
        // multiplication of a digit in num1.
        i_n1++;
    }
 
    // ignore '0's from the right
    int i = result.size() - 1;
    while (i>=0 && result[i] == 0)
    i--;
 
    // If all were '0's - means either both or
    // one of num1 or num2 were '0'
    if (i == -1)
    return "0";
 
    // generate the result string
    string s = "";
     
    while (i >= 0)
        s += std::to_string(result[i--]);
 
    return s;
}
 
// Driver code
int main()
{
    string str1 = "1235421415454545454545454544";
    string str2 = "1714546546546545454544548544544545";
     
    if((str1.at(0) == '-' || str2.at(0) == '-') &&
        (str1.at(0) != '-' || str2.at(0) != '-' ))
        cout<<"-";
 
 
    if(str1.at(0) == '-')
        str1 = str1.substr(1);
   
    if(str2.at(0) == '-')
        str2 = str2.substr(1);
 
    cout << multiply(str1, str2);
    return 0;
}

Java

// Java program to multiply two numbers
// represented as Strings.
class GFG
{
     
// Multiplies str1 and str2, and prints result.
static String multiply(String num1, String num2)
{
    int len1 = num1.length();
    int len2 = num2.length();
    if (len1 == 0 || len2 == 0)
        return "0";
 
    // will keep the result number in vector
    // in reverse order
    int result[] = new int[len1 + len2];
 
    // Below two indexes are used to
    // find positions in result.
    int i_n1 = 0;
    int i_n2 = 0;
     
    // Go from right to left in num1
    for (int i = len1 - 1; i >= 0; i--)
    {
        int carry = 0;
        int n1 = num1.charAt(i) - '0';
 
        // To shift position to left after every
        // multipliccharAtion of a digit in num2
        i_n2 = 0;
         
        // Go from right to left in num2            
        for (int j = len2 - 1; j >= 0; j--)
        {
            // Take current digit of second number
            int n2 = num2.charAt(j) - '0';
 
            // Multiply with current digit of first number
            // and add result to previously stored result
            // charAt current position.
            int sum = n1 * n2 + result[i_n1 + i_n2] + carry;
 
            // Carry for next itercharAtion
            carry = sum / 10;
 
            // Store result
            result[i_n1 + i_n2] = sum % 10;
 
            i_n2++;
        }
 
        // store carry in next cell
        if (carry > 0)
            result[i_n1 + i_n2] += carry;
 
        // To shift position to left after every
        // multipliccharAtion of a digit in num1.
        i_n1++;
    }
 
    // ignore '0's from the right
    int i = result.length - 1;
    while (i >= 0 && result[i] == 0)
    i--;
 
    // If all were '0's - means either both
    // or one of num1 or num2 were '0'
    if (i == -1)
    return "0";
 
    // genercharAte the result String
    String s = "";
     
    while (i >= 0)
        s += (result[i--]);
 
    return s;
}
 
// Driver code
public static void main(String[] args)
{
    String str1 = "1235421415454545454545454544";
    String str2 = "1714546546546545454544548544544545";
 
    if ((str1.charAt(0) == '-' || str2.charAt(0) == '-') &&
        (str1.charAt(0) != '-' || str2.charAt(0) != '-'))
        System.out.print("-");
 
    if (str1.charAt(0) == '-')
        str1 = str1.substring(1);
   
    if (str2.charAt(0) == '-')
        str2 = str2.substring(1);
 
    System.out.println(multiply(str1, str2));
}
}
 
// This code is contributed by ankush_953

Python3

# Python3 program to multiply two numbers
# represented as strings.
 
# Multiplies str1 and str2, and prints result.
def multiply(num1, num2):
    len1 = len(num1)
    len2 = len(num2)
    if len1 == 0 or len2 == 0:
        return "0"
 
    # will keep the result number in vector
    # in reverse order
    result = [0] * (len1 + len2)
     
    # Below two indexes are used to
    # find positions in result.
    i_n1 = 0
    i_n2 = 0
 
    # Go from right to left in num1
    for i in range(len1 - 1, -1, -1):
        carry = 0
        n1 = ord(num1[i]) - 48
 
        # To shift position to left after every
        # multiplication of a digit in num2
        i_n2 = 0
 
        # Go from right to left in num2
        for j in range(len2 - 1, -1, -1):
             
            # Take current digit of second number
            n2 = ord(num2[j]) - 48
         
            # Multiply with current digit of first number
            # and add result to previously stored result
            # at current position.
            summ = n1 * n2 + result[i_n1 + i_n2] + carry
 
            # Carry for next iteration
            carry = summ // 10
 
            # Store result
            result[i_n1 + i_n2] = summ % 10
 
            i_n2 += 1
 
            # store carry in next cell
        if (carry > 0):
            result[i_n1 + i_n2] += carry
 
            # To shift position to left after every
            # multiplication of a digit in num1.
        i_n1 += 1
         
        # print(result)
 
    # ignore '0's from the right
    i = len(result) - 1
    while (i >= 0 and result[i] == 0):
        i -= 1
 
    # If all were '0's - means either both or
    # one of num1 or num2 were '0'
    if (i == -1):
        return "0"
 
    # generate the result string
    s = ""
    while (i >= 0):
        s += chr(result[i] + 48)
        i -= 1
 
    return s
 
# Driver code
str1 = "1235421415454545454545454544"
str2 = "1714546546546545454544548544544545"
 
if((str1[0] == '-' or str2[0] == '-') and
   (str1[0] != '-' or str2[0] != '-')):
    print("-", end = '')
 
 
if(str1[0] == '-' and str2[0] != '-'):
    str1 = str1[1:]
elif(str1[0] != '-' and str2[0] == '-'):
    str2 = str2[1:]
elif(str1[0] == '-' and str2[0] == '-'):
    str1 = str1[1:]
    str2 = str2[1:]
print(multiply(str1, str2))
 
# This code is contributed by ankush_953

C#

// C# program to multiply two numbers
// represented as Strings.
using System;
 
class GFG
{
     
// Multiplies str1 and str2, and prints result.
static String multiply(String num1, String num2)
{
    int len1 = num1.Length;
    int len2 = num2.Length;
    if (len1 == 0 || len2 == 0)
        return "0";
 
    // will keep the result number in vector
    // in reverse order
    int []result = new int[len1 + len2];
 
    // Below two indexes are used to
    // find positions in result.
    int i_n1 = 0;
    int i_n2 = 0;
    int i;
     
    // Go from right to left in num1
    for (i = len1 - 1; i >= 0; i--)
    {
        int carry = 0;
        int n1 = num1[i] - '0';
 
        // To shift position to left after every
        // multipliccharAtion of a digit in num2
        i_n2 = 0;
         
        // Go from right to left in num2            
        for (int j = len2 - 1; j >= 0; j--)
        {
            // Take current digit of second number
            int n2 = num2[j] - '0';
 
            // Multiply with current digit of first number
            // and add result to previously stored result
            // charAt current position.
            int sum = n1 * n2 + result[i_n1 + i_n2] + carry;
 
            // Carry for next itercharAtion
            carry = sum / 10;
 
            // Store result
            result[i_n1 + i_n2] = sum % 10;
 
            i_n2++;
        }
 
        // store carry in next cell
        if (carry > 0)
            result[i_n1 + i_n2] += carry;
 
        // To shift position to left after every
        // multipliccharAtion of a digit in num1.
        i_n1++;
    }
 
    // ignore '0's from the right
    i = result.Length - 1;
    while (i >= 0 && result[i] == 0)
    i--;
 
    // If all were '0's - means either both
    // or one of num1 or num2 were '0'
    if (i == -1)
    return "0";
 
    // genercharAte the result String
    String s = "";
     
    while (i >= 0)
        s += (result[i--]);
 
    return s;
}
 
// Driver code
public static void Main(String[] args)
{
    String str1 = "1235421415454545454545454544";
    String str2 = "1714546546546545454544548544544545";
 
    if ((str1[0] == '-' || str2[0] == '-') &&
        (str1[0] != '-' || str2[0] != '-'))
        Console.Write("-");
 
    if (str1[0] == '-' && str2[0] != '-')
    {
        str1 = str1.Substring(1);
    }
    else if (str1[0] != '-' && str2[0] == '-')
    {
        str2 = str2.Substring(1);
    }
    else if (str1[0] == '-' && str2[0] == '-')
    {
        str1 = str1.Substring(1);
        str2 = str2.Substring(1);
    }
    Console.WriteLine(multiply(str1, str2));
}
}
 
// This code is contributed by Rajput-Ji

输出：

2118187521397235888154583183918321221520083884298838480662480

上面的代码改编自Gaurav提供的代码。
时间复杂度： O(m * n)，其中m和n是需要相乘的两个数字的长度。
另一种方法：

Java

// Java program to multiply two numbers represented
// as strings.
import java.util.Scanner;
 
public class StringMultiplication {
    // Driver code
    public static void main(String[] args)
    {
        String num1 = "1235421415454545454545454544";
        String tempnum1 = num1;
        String num2 = "1714546546546545454544548544544545";
        String tempnum2 = num2;
 
        // Check condition if one string is negative
        if (num1.charAt(0) == '-'
            && num2.charAt(0) != '-') {
            num1 = num1.substring(1);
        }
        else if (num1.charAt(0) != '-'
                 && num2.charAt(0) == '-') {
            num2 = num2.substring(1);
        }
        else if (num1.charAt(0) == '-'
                 && num2.charAt(0) == '-') {
            num1 = num1.substring(1);
            num2 = num2.substring(1);
        }
        String s1
            = new StringBuffer(num1).reverse().toString();
        String s2
            = new StringBuffer(num2).reverse().toString();
 
        int[] m = new int[s1.length() + s2.length()];
 
        // Go from right to left in num1
        for (int i = 0; i < s1.length(); i++) {
            // Go from right to left in num2
            for (int j = 0; j < s2.length(); j++) {
                m[i + j] = m[i + j]
                           + (s1.charAt(i) - '0')
                                 * (s2.charAt(j) - '0');
            }
        }
 
        String product = new String();
        // Multiply with current digit of first number
        // and add result to previously stored product
        // at current position.
        for (int i = 0; i < m.length; i++) {
            int digit = m[i] % 10;
            int carry = m[i] / 10;
            if (i + 1 < m.length) {
                m[i + 1] = m[i + 1] + carry;
            }
            product = digit + product;
        }
 
        // ignore '0's from the right
        while (product.length() > 1
               && product.charAt(0) == '0') {
            product = product.substring(1);
        }
 
        // Check condition if one string is negative
        if (tempnum1.charAt(0) == '-'
            && tempnum2.charAt(0) != '-') {
            product = new StringBuffer(product)
                          .insert(0, '-')
                          .toString();
        }
        else if (tempnum1.charAt(0) != '-'
                 && tempnum2.charAt(0) == '-') {
            product = new StringBuffer(product)
                          .insert(0, '-')
                          .toString();
        }
        else if (tempnum1.charAt(0) == '-'
                 && tempnum2.charAt(0) == '-') {
            product = product;
        }
        System.out.println("Product of the two numbers is :"
                           + "\n" + product);
    }
}

输出：

2118187521397235888154583183918321221520083884298838480662480