将表示为字符串的大数相乘

给定两个正数作为字符串。数字可能非常大(可能不适合长整型),任务是找到这两个数字的乘积。 例如:

null
Input : num1 = 4154          num2 = 51454Output : 213739916 Input :  num1 = 654154154151454545415415454           num2 = 63516561563156316545145146514654 Output : 41549622603955309777243716069997997007620439937711509062916

这个想法基于学校数学。

multiplication

我们从第二个数的最后一位开始,把它和第一个数相乘。然后我们将第二个数字的第二位与第一个数字相乘,依此类推。我们把所有这些乘法相加。在加法时,我们将第i次乘法移位。 下面的解决方案中使用的方法是只保留一个数组作为结果。我们在一个循环中遍历第一个和第二个数字的所有数字,并将结果添加到适当的位置。

C++

// C++ program to multiply two numbers represented
// as strings.
#include<bits/stdc++.h>
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< int > 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

以上代码改编自Gaurav提供的代码。 时间复杂性: O(m*n),其中m和n是需要相乘的两个数的长度。 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 :"
+ "" + product);
}
}


输出:

2118187521397235888154583183918321221520083884298838480662480

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