两个大数的和

给定两个数字作为字符串。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和。
例子：

Input  : str1 = "3333311111111111", 
         str2 =   "44422222221111"
Output : 3377733333332222

Input  : str1 = "7777555511111111", 
         str2 =    "3332222221111"
Output : 7780887733332222

这个想法是基于学校数学的。我们从头到尾遍历两个字符串，并逐个加一位，并跟踪进位。为了简化过程，我们执行以下操作：
1)颠倒两个字符串。
2)始终从第0个索引(在反向字符串)到较小字符串的末尾一一加数，将总和％10附加到结果末尾并跟踪进位为sum / 10。
3)最后反转结果。

C++

// C++ program to find sum of two large numbers.
#include
using namespace std;
 
// Function for finding sum of larger numbers
string findSum(string str1, string str2)
{
    // Before proceeding further, make sure length
    // of str2 is larger.
    if (str1.length() > str2.length())
        swap(str1, str2);
 
    // Take an empty string for storing result
    string str = "";
 
    // Calculate length of both string
    int n1 = str1.length(), n2 = str2.length();
 
    // Reverse both of strings
    reverse(str1.begin(), str1.end());
    reverse(str2.begin(), str2.end());
 
    int carry = 0;
    for (int i=0; i

---

Java

// Java program to find sum of two large numbers.
import java.util.*;
class GFG
{
// Function for finding sum of larger numbers
static String findSum(String str1, String str2)
{
    // Before proceeding further, make sure length
    // of str2 is larger.
    if (str1.length() > str2.length()){
        String t = str1;
        str1 = str2;
        str2 = t;
    }
 
    // Take an empty String for storing result
    String str = "";
 
    // Calculate length of both String
    int n1 = str1.length(), n2 = str2.length();
 
    // Reverse both of Strings
    str1=new StringBuilder(str1).reverse().toString();
    str2=new StringBuilder(str2).reverse().toString();
 
    int carry = 0;
    for (int i = 0; i < n1; i++)
    {
        // Do school mathematics, compute sum of
        // current digits and carry
        int sum = ((int)(str1.charAt(i) - '0') +
                    (int)(str2.charAt(i) - '0') + carry);
        str += (char)(sum % 10 + '0');
 
        // Calculate carry for next step
        carry = sum / 10;
    }
 
    // Add remaining digits of larger number
    for (int i = n1; i < n2; i++)
    {
        int sum = ((int)(str2.charAt(i) - '0') + carry);
        str += (char)(sum % 10 + '0');
        carry = sum / 10;
    }
 
    // Add remaining carry
    if (carry > 0)
        str += (char)(carry + '0');
 
    // reverse resultant String
    str = new StringBuilder(str).reverse().toString();
 
    return str;
}
 
// Driver code
public static void main(String[] args)
{
    String str1 = "12";
    String str2 = "198111";
    System.out.println(findSum(str1, str2));
}
}
// This code is contributed by mits

Python3

# Python3 program to find sum of
# two large numbers.
 
# Function for finding sum of
# larger numbers
def findSum(str1, str2):
     
    # Before proceeding further,
    # make sure length of str2 is larger.
    if (len(str1) > len(str2)):
        t = str1;
        str1 = str2;
        str2 = t;
 
    # Take an empty string for
    # storing result
    str = "";
 
    # Calculate length of both string
    n1 = len(str1);
    n2 = len(str2);
 
    # Reverse both of strings
    str1 = str1[::-1];
    str2 = str2[::-1];
 
    carry = 0;
    for i in range(n1):
         
        # Do school mathematics, compute
        # sum of current digits and carry
        sum = ((ord(str1[i]) - 48) +
              ((ord(str2[i]) - 48) + carry));
        str += chr(sum % 10 + 48);
 
        # Calculate carry for next step
        carry = int(sum / 10);
 
    # Add remaining digits of larger number
    for i in range(n1, n2):
        sum = ((ord(str2[i]) - 48) + carry);
        str += chr(sum % 10 + 48);
        carry = (int)(sum / 10);
 
    # Add remaining carry
    if (carry):
        str += chr(carry + 48);
 
    # reverse resultant string
    str = str[::-1];
 
    return str;
 
# Driver code
str1 = "12";
str2 = "198111";
print(findSum(str1, str2));
 
# This code is contributed by mits

C#

// C# program to find sum of two large numbers.
using System;
class GFG
{
// Function for finding sum of larger numbers
static string findSum(string str1, string str2)
{
    // Before proceeding further, make sure length
    // of str2 is larger.
    if (str1.Length > str2.Length){
        string t = str1;
        str1 = str2;
        str2 = t;
    }
 
    // Take an empty string for storing result
    string str = "";
 
    // Calculate length of both string
    int n1 = str1.Length, n2 = str2.Length;
 
    // Reverse both of strings
    char[] ch = str1.ToCharArray();
    Array.Reverse( ch );
    str1 = new string( ch );
    char[] ch1 = str2.ToCharArray();
    Array.Reverse( ch1 );
    str2 = new string( ch1 );
 
    int carry = 0;
    for (int i = 0; i < n1; i++)
    {
        // Do school mathematics, compute sum of
        // current digits and carry
        int sum = ((int)(str1[i] - '0') +
                (int)(str2[i] - '0') + carry);
        str += (char)(sum % 10 + '0');
 
        // Calculate carry for next step
        carry = sum/10;
    }
 
    // Add remaining digits of larger number
    for (int i = n1; i < n2; i++)
    {
        int sum = ((int)(str2[i] - '0') + carry);
        str += (char)(sum % 10 + '0');
        carry = sum/10;
    }
 
    // Add remaining carry
    if (carry > 0)
        str += (char)(carry + '0');
 
    // reverse resultant string
    char[] ch2 = str.ToCharArray();
    Array.Reverse( ch2 );
    str = new string( ch2 );
 
    return str;
}
 
// Driver code
static void Main()
{
    string str1 = "12";
    string str2 = "198111";
    Console.WriteLine(findSum(str1, str2));
}
}
 
// This code is contributed by mits

PHP

 strlen($str2)) {
        $t=$str1;
        $str1=$str2;
        $str2=$t;
    }
 
    // Take an empty string for storing result
    $str = "";
 
    // Calculate length of both string
    $n1 = strlen($str1);
    $n2 = strlen($str2);
 
    // Reverse both of strings
    $str1 = strrev($str1);
    $str2 = strrev($str2);
 
    $carry = 0;
    for ($i=0; $i<$n1; $i++)
    {
        // Do school mathematics, compute sum of
        // current digits and carry
        $sum = ((ord($str1[$i])-48)+((ord($str2[$i])-48)+$carry));
        $str.=chr($sum%10 + 48);
 
        // Calculate carry for next step
        $carry = (int)($sum/10);
    }
 
    // Add remaining digits of larger number
    for ($i=$n1; $i<$n2; $i++)
    {
        $sum = ((ord($str2[$i])-48)+$carry);
        $str.=chr($sum%10 + 48);
        $carry = (int)($sum/10);
    }
 
    // Add remaining carry
    if ($carry)
        $str.=chr($carry+48);
 
    // reverse resultant string
    $str=strrev($str);
 
    return $str;
}
 
// Driver code
  
    $str1 = "12";
    $str2 = "198111";
    echo findSum($str1, $str2);
 
// This code is contributed by mits
?>

输出：

198123

优化：
我们可以通过从头开始遍历它们来避免前两个字符串反向操作。下面是优化的解决方案。

C++

// C++ program to find sum of two large numbers.
#include
using namespace std;
 
// Function for finding sum of larger numbers
string findSum(string str1, string str2)
{
    // Before proceeding further, make sure length
    // of str2 is larger.
    if (str1.length() > str2.length())
        swap(str1, str2);
 
    // Take an empty string for storing result
    string str = "";
 
    // Calculate length of both string
    int n1 = str1.length(), n2 = str2.length();
    int diff = n2 - n1;
 
    // Initially take carry zero
    int carry = 0;
 
    // Traverse from end of both strings
    for (int i=n1-1; i>=0; i--)
    {
        // Do school mathematics, compute sum of
        // current digits and carry
        int sum = ((str1[i]-'0') +
                   (str2[i+diff]-'0') +
                   carry);
        str.push_back(sum%10 + '0');
        carry = sum/10;
    }
 
    // Add remaining digits of str2[]
    for (int i=n2-n1-1; i>=0; i--)
    {
        int sum = ((str2[i]-'0')+carry);
        str.push_back(sum%10 + '0');
        carry = sum/10;
    }
 
    // Add remaining carry
    if (carry)
        str.push_back(carry+'0');
 
    // reverse resultant string
    reverse(str.begin(), str.end());
 
    return str;
}
 
// Driver code
int main()
{
    string str1 = "12";
    string str2 = "198111";
    cout << findSum(str1, str2);
    return 0;
}

Java

// Java program to find sum of two large numbers.
import java.util.*;
 
class GFG{
     
 // Function for finding sum of larger numbers
static String findSum(String str1, String str2)
{
    // Before proceeding further, make sure length
    // of str2 is larger.
    if (str1.length() > str2.length()){
        String t = str1;
        str1 = str2;
        str2 = t;
    }
 
    // Take an empty String for storing result
    String str = "";
 
    // Calculate length of both String
    int n1 = str1.length(), n2 = str2.length();
    int diff = n2 - n1;
 
    // Initially take carry zero
    int carry = 0;
 
    // Traverse from end of both Strings
    for (int i = n1 - 1; i>=0; i--)
    {
        // Do school mathematics, compute sum of
        // current digits and carry
        int sum = ((int)(str1.charAt(i)-'0') +
            (int)(str2.charAt(i+diff)-'0') + carry);
        str += (char)(sum % 10 + '0');
        carry = sum / 10;
    }
 
    // Add remaining digits of str2[]
    for (int i = n2 - n1 - 1; i >= 0; i--)
    {
        int sum = ((int)(str2.charAt(i) - '0') + carry);
        str += (char)(sum % 10 + '0');
        carry = sum / 10;
    }
 
    // Add remaining carry
    if (carry > 0)
        str += (char)(carry + '0');
 
    // reverse resultant String
    return new StringBuilder(str).reverse().toString();
}
 
// Driver code
public static void main(String[] args)
{
    String str1 = "12";
    String str2 = "198111";
    System.out.println(findSum(str1, str2));
}
}
 
// This code is contributed by mits

的Python 3

# python 3 program to find sum of two large numbers.
  
# Function for finding sum of larger numbers
def findSum(str1, str2):
 
    # Before proceeding further, make sure length
    # of str2 is larger.
    if len(str1)> len(str2):
        temp = str1
        str1 = str2
        str2 = temp
  
    # Take an empty string for storing result
    str3 = ""
  
    # Calculate length of both string
    n1 = len(str1)
    n2 = len(str2)
    diff = n2 - n1
  
    # Initially take carry zero
    carry = 0
  
    # Traverse from end of both strings
    for i in range(n1-1,-1,-1):
     
        # Do school mathematics, compute sum of
        # current digits and carry
       
        sum = ((ord(str1[i])-ord('0')) +
                   int((ord(str2[i+diff])-ord('0'))) + carry)
      
        str3 = str3+str(sum%10 )
         
        
        carry = sum//10
  
    # Add remaining digits of str2[]
    for i in range(n2-n1-1,-1,-1):
     
        sum = ((ord(str2[i])-ord('0'))+carry)
        str3 = str3+str(sum%10 )
        carry = sum//10
  
    # Add remaining carry
    if (carry):
        str3+str(carry+'0')
  
    # reverse resultant string
    str3 = str3[::-1]
  
    return str3
  
# Driver code
if __name__ == "__main__":
    str1 = "12"
    str2 = "198111"
    print(findSum(str1, str2))
 
# This code is contributed by ChitraNayal

C＃

// C# program to find sum of two large numbers.
using System;
 
class GFG{
     
// Function for finding sum of larger numbers
static string findSum(string str1, string str2)
{
    // Before proceeding further, make sure length
    // of str2 is larger.
    if (str1.Length > str2.Length)
    {
        string t = str1;
        str1 = str2;
        str2 = t;
    }
 
    // Take an empty string for storing result
    string str = "";
 
    // Calculate length of both string
    int n1 = str1.Length, n2 = str2.Length;
    int diff = n2 - n1;
 
    // Initially take carry zero
    int carry = 0;
 
    // Traverse from end of both strings
    for (int i = n1 - 1; i >= 0; i--)
    {
        // Do school mathematics, compute sum of
        // current digits and carry
        int sum = ((int)(str1[i] - '0') +
                (int)(str2[i + diff]-'0') + carry);
        str += (char)(sum % 10 + '0');
        carry = sum / 10;
    }
 
    // Add remaining digits of str2[]
    for (int i = n2 - n1 - 1; i >= 0; i--)
    {
        int sum = ((int)(str2[i] - '0') + carry);
        str += (char)(sum % 10 + '0');
        carry = sum / 10;
    }
 
    // Add remaining carry
    if (carry > 0)
        str += (char)(carry + '0');
 
    // reverse resultant string
    char[] ch2 = str.ToCharArray();
    Array.Reverse(ch2);
    return new string(ch2);
}
 
// Driver code
static void Main()
{
    string str1 = "12";
    string str2 = "198111";
    Console.WriteLine(findSum(str1, str2));
}
}
 
// This code is contributed by mits

的PHP

 strlen($str2))
    {
        $temp = $str1;
        $str1 = $str2;
        $str2 = $temp;
    }
 
    // Take an empty string for storing result
    $str3 = "";
 
    // Calculate length of both string
    $n1 = strlen($str1);
    $n2 = strlen($str2);
    $diff = $n2 - $n1;
 
    // Initially take carry zero
    $carry = 0;
 
    // Traverse from end of both strings
    for ($i = $n1 - 1; $i >= 0; $i--)
    {
        // Do school mathematics, compute sum 
        // of current digits and carry
        $sum = ((ord($str1[$i]) - ord('0')) +
               ((ord($str2[$i + $diff]) -
                 ord('0'))) + $carry);
     
        $str3 .= chr($sum % 10 + ord('0'));
         
         
        $carry = (int)($sum / 10);
    }
 
    // Add remaining digits of str2[]
    for ($i = $n2 - $n1 - 1; $i >= 0; $i--)
    {
        $sum = ((ord($str2[$i]) - ord('0')) + $carry);
        $str3 .= chr($sum % 10 + ord('0'));
        $carry = (int)($sum / 10);
    }
 
    // Add remaining carry
    if ($carry)
        $str3 .= chr($carry + ord('0'));
 
    // reverse resultant string
    return strrev($str3);
}
 
// Driver code
$str1 = "12";
$str2 = "198111";
print(findSum($str1, $str2));
 
// This code is contributed by mits
?>

输出：

198123

时间复杂度： O(n1 + n2)，其中n1和n2是表示数字的两个输入字符串的长度。