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Calcular la suma de todos los números presentes en una cadena.

Pruébalo en GfG Practice ' title= #practiceLinkDiv { mostrar: ninguno !importante; }

dada una cuerda S que contiene caracteres alfanuméricos La tarea consiste en calcular la suma de todos los números presentes en la cadena.

Ejemplos:  



Aporte:  1abc23
Producción: 24
Explicación: 1 + 23 = 24

Aporte:  frikis4geeks
Producción: 4

Aporte:  1abc2x30yz67
Producción: 100



Práctica recomendada Suma de números en cadena ¡Pruébalo!

Acercarse:

Escanee cada carácter de la cadena de entrada y si un número está formado por caracteres consecutivos de la cadena, incremente el resultado por esa cantidad. La única parte complicada de esta pregunta es que varios dígitos consecutivos se consideran un número.

Siga los pasos a continuación para implementar la idea:

  • Crea una cadena vacía temperatura y un numero entero suma .
  • Iterar sobre todos los caracteres de la cadena.
    • Si el carácter es un dígito numérico, agréguelo a temperatura .
    • De lo contrario, convierta la cadena temporal en un número y agréguela a suma vacío temperatura .
  • Devuelve suma + número obtenido de temp.

A continuación se muestra la implementación del enfoque anterior:



C++
// C++ program to calculate sum of all numbers present // in a string containing alphanumeric characters #include    using namespace std; // Function to calculate sum of all numbers present // in a string containing alphanumeric characters int findSum(string str) {  // A temporary string  string temp = '';  // holds sum of all numbers present in the string  int sum = 0;  // read each character in input string  for (char ch : str) {  // if current character is a digit  if (isdigit(ch))  temp += ch;  // if current character is an alphabet  else {  // increment sum by number found earlier  // (if any)  sum += atoi(temp.c_str());  // reset temporary string to empty  temp = '';  }  }  // atoi(temp.c_str()) takes care of trailing  // numbers  return sum + atoi(temp.c_str()); } // Driver code int main() {  // input alphanumeric string  string str = '12abc20yz68';  // Function call  cout << findSum(str);  return 0; } 
Java
// Java program to calculate sum of all numbers present // in a string containing alphanumeric characters import java.io.*; class GFG {  // Function to calculate sum of all numbers present  // in a string containing alphanumeric characters  static int findSum(String str)  {  // A temporary string  String temp = '0';  // holds sum of all numbers present in the string  int sum = 0;  // read each character in input string  for (int i = 0; i < str.length(); i++) {  char ch = str.charAt(i);  // if current character is a digit  if (Character.isDigit(ch))  temp += ch;  // if current character is an alphabet  else {  // increment sum by number found earlier  // (if any)  sum += Integer.parseInt(temp);  // reset temporary string to empty  temp = '0';  }  }  // atoi(temp.c_str()) takes care of trailing  // numbers  return sum + Integer.parseInt(temp);  }  // Driver code  public static void main(String[] args)  {  // input alphanumeric string  String str = '12abc20yz68';  // Function call  System.out.println(findSum(str));  } } // This code is contributed by AnkitRai01 
Python3
# Python3 program to calculate sum of # all numbers present in a string # containing alphanumeric characters # Function to calculate sum of all # numbers present in a string # containing alphanumeric characters def findSum(str1): # A temporary string temp = '0' # holds sum of all numbers # present in the string Sum = 0 # read each character in input string for ch in str1: # if current character is a digit if (ch.isdigit()): temp += ch # if current character is an alphabet else: # increment Sum by number found # earlier(if any) Sum += int(temp) # reset temporary string to empty temp = '0' # atoi(temp.c_str1()) takes care # of trailing numbers return Sum + int(temp) # Driver code # input alphanumeric string str1 = '12abc20yz68' # Function call print(findSum(str1)) # This code is contributed # by mohit kumar 
C#
// C# program to calculate sum of // all numbers present in a string // containing alphanumeric characters using System; class GFG {  // Function to calculate sum of  // all numbers present in a string  // containing alphanumeric characters  static int findSum(String str)  {  // A temporary string  String temp = '0';  // holds sum of all numbers  // present in the string  int sum = 0;  // read each character in input string  for (int i = 0; i < str.Length; i++) {  char ch = str[i];  // if current character is a digit  if (char.IsDigit(ch))  temp += ch;  // if current character is an alphabet  else {  // increment sum by number found earlier  // (if any)  sum += int.Parse(temp);  // reset temporary string to empty  temp = '0';  }  }  // atoi(temp.c_str()) takes care of trailing  // numbers  return sum + int.Parse(temp);  }  // Driver code  public static void Main(String[] args)  {  // input alphanumeric string  String str = '12abc20yz68';  // Function call  Console.WriteLine(findSum(str));  } } // This code is contributed by PrinciRaj1992 
JavaScript
<script> // Javascript program to calculate // sum of all numbers present // in a string containing  // alphanumeric characters    // Function to calculate sum   // of all numbers present  // in a string containing   // alphanumeric characters  function findSum(str)  {  // A temporary string  let temp = '0';    // holds sum of all numbers   // present in the string  let sum = 0;    // read each character in input string  for (let i = 0; i < str.length; i++) {  let ch = str[i];    // if current character is a digit  if (!isNaN(String(ch) * 1))  temp += ch;    // if current character is an alphabet  else {  // increment sum by number found earlier  // (if any)  sum += parseInt(temp);    // reset temporary string to empty  temp = '0';  }  }    // atoi(temp.c_str()) takes care of trailing  // numbers  return sum + parseInt(temp);  }    // Driver code  // input alphanumeric string  let str = '12abc20yz68';    // Function call  document.write(findSum(str));   // This code is contributed by unknown2108 </script> 

Producción
100


Complejidad del tiempo: O(N) donde n es la longitud de la cuerda. 
Espacio Auxiliar: O(N) donde n es la longitud de la cuerda.

Calcule la suma de todos los números presentes en una cadena usando recursividad

La idea es recorrer recursivamente la cadena y descubrir el números Luego suma estos números a resultado por fin regresa el resultado

Siga los pasos a continuación para implementar la idea:

  • Crea una cadena vacía temperatura y un numero entero suma .
  • Recorre recursivamente los caracteres de cada índice. i de a longitud - 1 .
    • Si yo = N-1 luego verifique si el carácter actual es un dígito cadena[i] - '0' .
    • De lo contrario regresar .
    • Si cadena [i] es un dígito
      • Ejecute un bucle for con contador de i a norte-1 .
        • Si el carácter es un dígito numérico, agréguelo a temperatura .
        • De lo contrario, rompe.
      • Devolver suma del valor numérico de temp + repetir para el índice j .

A continuación se muestra la implementación del enfoque anterior:

C++
// C++ program to calculate sum of all numbers // present in a string containing alphanumeric // characters #include    using namespace std; int solve(string& str int i int n) {  // if string is empty  if (i >= n)  return 0;  // if on the last index  if (i == n - 1) {  // if last digit is numeric  if (isdigit(str[i])) {  return str[i] - '0';  }  else {  return 0;  }  }  // if current char is digit  // then sum the consecutive digits  if (isdigit(str[i])) {  // declared an empty string  string temp = '';  int j;  // start from that index  // sum all the consecutive digits  for (j = i; j < n; j++) {  // if current char is digit  // add it to the temp string  if (isdigit(str[j]))  temp += str[j];  // if it is not a digit  // break instantly  else  break;  }  // add the number associated to temp  // with the answer recursion will bring  return stoi(temp) + solve(str j n);  }  // else call from the next index  else {  solve(str i + 1 n);  } } int findSum(string str) {  // recursiven function  return solve(str 0 str.size()); } // Driver code int main() {  // input alphanumeric string  string str = '12abc20yz68';  // Function call  cout << findSum(str);  return 0; } 
Java
import java.util.Scanner; class Main {  static int solve(String str int i int n) {  // if string is empty  if (i >= n)  return 0;  // if on the last index  if (i == n - 1) {  // if last digit is numeric  if (Character.isDigit(str.charAt(i))) {  return str.charAt(i) - '0';  }  else {  return 0;  }  }  // if current char is digit  // then sum the consecutive digits  if (Character.isDigit(str.charAt(i))) {  // declared an empty string  String temp = '';  int j;  // start from that index  // sum all the consecutive digits  for (j = i; j < n; j++) {  // if current char is digit  // add it to the temp string  if (Character.isDigit(str.charAt(j)))  temp += str.charAt(j);  // if it is not a digit  // break instantly  else  break;  }  // add the number associated to temp  // with the answer recursion will bring  return Integer.parseInt(temp) + solve(str j n);  }  // else call from the next index  else {  return solve(str i + 1 n);  }  }  static int findSum(String str) {  // recursiven function  return solve(str 0 str.length());  }  // Driver code  public static void main(String[] args) {  // input alphanumeric string  String str = '12abc20yz68';  // Function call  System.out.println(findSum(str));  } } // This code contributed by Ajax 
Python3
def findSum(str): # variable to store sum result = 0 temp = '' for i in range(len(str)): if str[i].isnumeric(): temp += str[i] if i == len(str) - 1: result += int(temp) else: if temp != '': result += int(temp) temp = '' return result # driver code if __name__ == '__main__': # input alphanumeric string str = '12abc20yz68' print(findSum(str)) #This code contributed by Shivam Tiwari 
C#
// C# program to calculate sum of all numbers // present in a string containing alphanumeric // characters using System; using System.Linq; using System.Collections.Generic; class GFG  {  static bool isdigit(char c)  {  if(c>='0' && c<='9')  return true;  return false;  }  static int solve(string str int i int n)  {  // if string is empty  if (i >= n)  return 0;    // if on the last index  if (i == n - 1) {    // if last digit is numeric  if (isdigit(str[i])) {  return str[i];  }  else {  return 0;  }  }    // if current char is digit  // then sum the consecutive digits  if (isdigit(str[i])) {    // declared an empty string  string temp = '';  int j;    // start from that index  // sum all the consecutive digits  for (j = i; j < n; j++) {    // if current char is digit  // add it to the temp string  if (isdigit(str[j]))  temp += str[j];    // if it is not a digit  // break instantly  else  break;  }    // add the number associated to temp  // with the answer recursion will bring  return Int32.Parse(temp) + solve(str j n);  }    // else call from the next index  else {  return solve(str i + 1 n);  }  }    static int findSum(string str)  {  // recursiven function  return solve(str 0 str.Length);  }    // Driver code  static public void Main()  {  // input alphanumeric string  string str = '12abc20yz68';    // Function call  Console.Write(findSum(str));    } } 
JavaScript
function findSum(str) {  // variable to store sum  let result = 0;  let temp = '';    for (let i = 0; i < str.length; i++) {  if (!isNaN(str[i])) {  temp += str[i];  if (i === str.length - 1) {  result += parseInt(temp);  }  } else {  if (temp !== '') {  result += parseInt(temp);  temp = '';  }  }  }  return result; } // driver code console.log(findSum('12abc20yz68')); // This code is contributed by Shivam Tiwari 

Producción
100

Complejidad del tiempo: EN) donde N es el tamaño de la cadena dada.
Espacio Auxiliar: EN) en el peor de los casos, puede costar O(N) llamadas recursivas

Calcule la suma de todos los números presentes en una cadena usando Regex en Python:

La idea es utilizar la función incorporada. Expresión regular de Python . 

A continuación se muestra la implementación del enfoque anterior:

C++14
#include    #include  // Function to calculate sum of all // numbers present in a string // containing alphanumeric characters int findSum(std::string str) {  // Regular Expression that matches  // digits in between a string  std::regex pattern('\d+');  std::smatch match;  int sum = 0;  while (std::regex_search(str match pattern)) {  sum += stoi(match[0].str());  str = match.suffix().str();  }  return sum; } // Driver code int main() {  // input alphanumeric string  std::string str = '12abc20yz68';  // Function call  std::cout << findSum(str) << std::endl;  return 0; } // This code is contributed by Shivam Tiwari 
Python3
# Python3 program to calculate sum of # all numbers present in a string # containing alphanumeric characters # Function to calculate sum of all # numbers present in a string # containing alphanumeric characters import re def find_sum(str1): # Regular Expression that matches # digits in between a string return sum(map(int re.findall('d+' str1))) # Driver code # input alphanumeric string str1 = '12abc20yz68' # Function call print(find_sum(str1)) # This code is contributed # by Venkata Ramana B 
JavaScript
// JavaScript program to calculate sum of // all numbers present in a string // containing alphanumeric characters // Function to calculate sum of all // numbers present in a string // containing alphanumeric characters function find_sum(str1) {  // Regular Expression that matches  // digits in between a string  return str1.match(/d+/g).reduce((acc val) => acc + parseInt(val) 0); } // Driver code // input alphanumeric string const str1 = '12abc20yz68'; // Function call console.log(find_sum(str1)); 
Java
import java.util.regex.*; public class Main {  // Function to calculate sum of all  // numbers present in a string  // containing alphanumeric characters  public static int findSum(String str)  {  // Regular Expression that matches  // digits in between a string  Pattern pattern = Pattern.compile('\d+');  Matcher matcher = pattern.matcher(str);  int sum = 0;  while (matcher.find()) {  sum += Integer.parseInt(matcher.group());  str = matcher.replaceFirst('');  matcher = pattern.matcher(str);  }  return sum;  }  // Driver code  public static void main(String[] args)  {  // input alphanumeric string  String str = '12abc20yz68';  // Function call  System.out.println(findSum(str));  } } 
C#
using System; using System.Text.RegularExpressions; public class GFG {  // Function to calculate sum of all  // numbers present in a string  // containing alphanumeric characters  public static int FindSum(string str)  {  // Regular Expression that matches  // digits in between a string  Regex pattern = new Regex(@'d+');  Match matcher = pattern.Match(str);  int sum = 0;  while (matcher.Success)  {  sum += Int32.Parse(matcher.Value);  str = pattern.Replace(str '' 1 matcher.Index);  matcher = pattern.Match(str);  }  return sum;  }  // Main method  static public void Main()  {  // input alphanumeric string  string str = '12abc20yz68';  // Function call  Console.WriteLine(FindSum(str));  } } 

Producción
100

Complejidad del tiempo: O(n) donde n es la longitud de la cuerda. 
Espacio Auxiliar: O(n) donde n es la longitud de la cuerda.