#practiceLinkDiv { display: none !important; }Dota virkne S kas satur burtciparu rakstzīmes Uzdevums ir aprēķināt visu virknē esošo skaitļu summu.
Piemēri:
Ieteicamā prakse Skaitļu summa virknē Izmēģiniet to!Ievade: 1abc23
Izvade: 24
Paskaidrojums: 1 + 23 = 24Ievade: geeks4geeks
Izvade: 4Ievade: 1abc2x30yz67
Izvade: 100
Pieeja:
Skenējiet katru ievades virknes rakstzīmi un, ja skaitli veido secīgas virknes rakstzīmes, palieliniet rezultāts par šo summu. Vienīgā šī jautājuma sarežģītā daļa ir tā, ka vairāki cipari pēc kārtas tiek uzskatīti par vienu skaitli.
Lai īstenotu ideju, veiciet tālāk norādītās darbības.
- Izveidojiet tukšu virkni temp un vesels skaitlis summa .
- Atkārtojiet visas virknes rakstzīmes.
- Ja rakstzīme ir cipars, pievienojiet to temp .
- Citādi konvertējiet temp virkni par skaitli un pievienojiet to summa tukšs temp .
- Atgriešanās summa + skaitlis, kas iegūts no temp.
Tālāk ir aprakstīta iepriekš minētās pieejas īstenošana.
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>
Izvade
100
Laika sarežģītība: O(N) kur n ir virknes garums.
Palīgtelpa: O(N) kur n ir virknes garums.
Aprēķiniet visu virknē esošo skaitļu summu, izmantojot rekursija
Ideja ir rekursīvi šķērsot virkni un noskaidrot cipariem pēc tam pievienojiet šos skaitļus rezultāts beidzot atgriezieties rezultāts .
Lai īstenotu ideju, veiciet tālāk norādītās darbības.
- Izveidojiet tukšu virkni temp un vesels skaitlis summa .
- Rekursīvi šķērsojiet katra indeksa rakstzīmes i no uz garums - 1 .
- Ja i = N-1 pēc tam pārbaudiet, vai pašreizējā rakstzīme ir ciparu atgriešana str[i] - '0' .
- Citādi atgriešanās .
- Ja str[i] ir cipars.
- Palaidiet for cilpu ar skaitītāju j no i uz N-1 .
- Ja rakstzīme ir cipars, pievienojiet to temp .
- Citādi pārtraukums.
- Atgriezties summa no temp + atkārtotas indeksa skaitliskās vērtības j .
- Palaidiet for cilpu ar skaitītāju j no i uz N-1 .
Tālāk ir aprakstīta iepriekš minētās pieejas īstenošana.
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
Izvade
100
Laika sarežģītība: O(N) kur N ir dotās virknes lielums.
Palīgtelpa: O(N) sliktākajā gadījumā tas var maksāt O(N) rekursīvus zvanus
Aprēķiniet visu virknē esošo skaitļu summu, izmantojot Regex programmā Python:
Ideja ir izmantot iebūvēto funkciju Python RegEx .
Zemāk ir aprakstīta iepriekš minētās pieejas īstenošana:
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)); } }
Izvade
100
Laika sarežģītība: O(n) kur n ir virknes garums.
Palīgtelpa: O(n) kur n ir virknes garums.