Apskaičiuokite visų skaičių, esančių eilutėje, sumą
Išbandykite GfG praktikoje 
#practiceLinkDiv { display: none !important; }
Išvestis
Išvestis
Išvestis
#practiceLinkDiv { display: none !important; } Duota eilutė S su raidiniais ir skaitiniais simboliais Užduotis yra apskaičiuoti visų eilutėje esančių skaičių sumą.
Pavyzdžiai:
Rekomenduojama praktika Skaičių suma eilutėje Išbandykite!Įvestis: 1abc23
Išvestis: 24
Paaiškinimas: 1 + 23 = 24Įvestis: geeks4geeks
Išvestis: 4Įvestis: 1abc2x30yz67
Išvestis: 100
Prieiga:
Nuskaitykite kiekvieną įvesties eilutės simbolį ir, jei skaičių sudaro iš eilės einantys simboliai, padidinkite rezultatas ta suma. Vienintelė sudėtinga šio klausimo dalis yra ta, kad keli skaitmenys iš eilės laikomi vienu skaičiumi.
Norėdami įgyvendinti idėją, atlikite šiuos veiksmus:
- Sukurkite tuščią eilutę temp ir sveikasis skaičius suma .
- Pakartokite visus eilutės simbolius.
- Jei simbolis yra skaitmuo, pridėkite jį prie temp .
- Kitu atveju konvertuokite temp eilutę į skaičių ir pridėkite prie suma tuščias temp .
- Grąžinama suma + skaičius, gautas iš temp.
Žemiau pateikiamas pirmiau minėto metodo įgyvendinimas:
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>
Išvestis
100
Laiko sudėtingumas: O(N) kur n yra eilutės ilgis.
Pagalbinė erdvė: O(N) kur n yra eilutės ilgis.
Apskaičiuokite visų skaičių, esančių eilutėje, sumą naudodami rekursija
Idėja yra rekursyviai pereiti per eilutę ir išsiaiškinti numeriai tada pridėkite šiuos skaičius prie rezultatas pagaliau grąžinti rezultatas .
Norėdami įgyvendinti idėją, atlikite šiuos veiksmus:
- Sukurkite tuščią eilutę temp ir sveikasis skaičius suma .
- Rekursyviai pereikite kiekvieno indekso simbolius i iš į ilgis - 1 .
- Jeigu i = N-1 tada patikrinkite, ar dabartinis simbolis yra skaitmens grąžinimas str[i] – '0' .
- Kitaip grįžti .
- Jeigu str[i] yra skaitmuo.
- Paleiskite for kilpą su skaitikliu j iš i į N-1 .
- Jei simbolis yra skaitmuo, pridėkite jį prie temp .
- Kitaip pertrauka.
- Grįžti suma skaitinės reikšmės temp + pasikartojimas indeksui j .
- Paleiskite for kilpą su skaitikliu j iš i į N-1 .
Žemiau pateikiamas pirmiau minėto metodo įgyvendinimas:
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
Išvestis
100
Laiko sudėtingumas: O(N) kur N yra nurodytos eilutės dydis.
Pagalbinė erdvė: O(N) blogiausiu atveju tai gali kainuoti O(N) rekursinių skambučių
Apskaičiuokite visų skaičių, esančių eilutėje, sumą naudodami Regex programoje Python:
Idėja yra naudoti įmontuotą funkciją Python RegEx .
Žemiau pateikiamas aukščiau pateikto požiūrio įgyvendinimas:
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 )); } }
Išvestis
100
Laiko sudėtingumas: O(n) kur n yra eilutės ilgis.
Pagalbinė erdvė: O(n) kur n yra eilutės ilgis.
