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:
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Input : num1 = 4154 num2 = 51454Output : 213739916 Input : num1 = 654154154151454545415415454 num2 = 63516561563156316545145146514654 Output : 41549622603955309777243716069997997007620439937711509062916
这个想法基于学校数学。
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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