Jei gautas teigiamas sveikasis skaičius, parašykite funkciją, kuri grąžina teisingą skaičių, jei nurodytas skaičius yra palindromas, kitu atveju klaidingas. Pavyzdžiui, 12321 yra palindromas, bet 1451 nėra palindromas.
Rekomenduojamos praktikos skaitmenų suma yra Pallindromas, ar ne Išbandykite!
1 būdas:
Tegul nurodytas skaičius yra ant vieno . Paprastas būdas išspręsti šią problemą yra pirmiausia atvirkštiniai skaitmenys ant vieno , tada palyginkite atvirkščiai ant vieno su ant vieno . Jei abu yra vienodi, grąžinkite teisingą, kitu atveju klaidingą.
Toliau pateikiamas įdomus metodas, įkvėptas 2 metodo tai paštu. Idėja yra sukurti kopiją ant vieno ir rekursyviai perduoti kopiją pagal nuorodą ir perduoti ant vieno pagal vertę. Rekursyviniuose skambučiuose padalinkite ant vieno 10 judant žemyn rekursijos medžiu. Judėdami aukštyn rekursijos medyje, padalykite kopiją iš 10. Kai jie susitinka funkcijoje, kuriai baigiasi visi antriniai skambučiai, paskutinis ant vieno bus i-asis skaitmuo nuo pradžios, o paskutinis kopijos skaitmuo bus i-asis skaitmuo nuo pabaigos.
C++
// A recursive C++ program to check> // whether a given number> // is palindrome or not> #include> using> namespace> std;> > // A function that returns true only> // if num contains one> // digit> int> oneDigit(>int> num)> {> > >// Comparison operation is faster> >// than division> >// operation. So using following> >// instead of 'return num> >// / 10 == 0;'> >return> (num>= 0 && skaičius<10);> }> > // A recursive function to find> // out whether num is> // palindrome or not. Initially, dupNum> // contains address of> // a copy of num.> bool> isPalUtil(>int> num,>int>* dupNum)> {> > >// Base case (needed for recursion> >// termination): This> >// statement mainly compares the> >// first digit with the> >// last digit> >if> (oneDigit(num))> >return> (num == (*dupNum) % 10);> > >// This is the key line in this> >// method. Note that all> >// recursive calls have a separate> >// copy of num, but they> >// all share same copy of *dupNum.> >// We divide num while> >// moving up the recursion tree> >if> (!isPalUtil(num / 10, dupNum))> >return> false>;> > >// The following statements are> >// executed when we move up> >// the recursion call tree> >*dupNum /= 10;> > >// At this point, if num%10 contains> >// i'th digit from> >// beginning, then (*dupNum)%10> >// contains i'th digit> >// from end> >return> (num % 10 == (*dupNum) % 10);> }> > // The main function that uses> // recursive function> // isPalUtil() to find out whether> // num is palindrome or not> int> isPal(>int> num)> {> > >// Check if num is negative,> >// make it positive> >if> (num <0)> >num = -num;> > >// Create a separate copy of num,> >// so that modifications> >// made to address dupNum don't> >// change the input number.> >// *dupNum = num> >int>* dupNum =>new> int>(num);> > >return> isPalUtil(num, dupNum);> }> > // Driver program to test> // above functions> int> main()> {> >int> n = 12321;> >isPal(n) ? cout <<>'Yes
'>: cout <<>'No'> << endl;> > >n = 12;> >isPal(n) ? cout <<>'Yes
'>: cout <<>'No'> << endl;> > >n = 88;> >isPal(n) ? cout <<>'Yes
'>: cout <<>'No'> << endl;> > >n = 8999;> >isPal(n) ? cout <<>'Yes
'>: cout <<>'No'>;> >return> 0;> }> > // this code is contributed by shivanisinghss2110> |
>
>
C
#include> #include> > // A function that returns true only> // if num contains one digit> int> oneDigit(>int> num)> {> >// Comparison operation is faster> >// than division operation.> >// So using the following instead of 'return num / 10 == 0;'> >return> (num>= 0 && skaičius<10);> }> > // A recursive function to find out whether> // num is palindrome or not.> // Initially, dupNum contains the address of a copy of num.> bool> isPalUtil(>int> num,>int>* dupNum)> {> >// Base case (needed for recursion termination):> >// This statement mainly compares the first digit with the last digit.> >if> (oneDigit(num))> >return> (num == (*dupNum) % 10);> > >// This is the key line in this method.> >// Note that all recursive calls have a separate copy of num,> >// but they all share the same copy of *dupNum.> >// We divide num while moving up the recursion tree.> >if> (!isPalUtil(num / 10, dupNum))> >return> false>;> > >// The following statements are executed when we move up the recursion call tree.> >*dupNum /= 10;> > >// At this point, if num % 10 contains the i'th digit from the beginning,> >// then (*dupNum) % 10 contains the i'th digit from the end.> >return> (num % 10 == (*dupNum) % 10);> }> > // The main function that uses the recursive function> // isPalUtil() to find out whether num is palindrome or not.> bool> isPal(>int> num)> {> >// Check if num is negative, make it positive.> >if> (num <0)> >num = -num;> > >// Create a separate copy of num, so that modifications> >// made to the address dupNum don't change the input number.> >int> dupNum = num;> > >return> isPalUtil(num, &dupNum);> }> > // Driver program to test above functions> int> main()> {> >int> n = 12321;> >isPal(n) ?>printf>(>'Yes
'>) :>printf>(>'No
'>);> > >n = 12;> >isPal(n) ?>printf>(>'Yes
'>) :>printf>(>'No
'>);> > >n = 88;> >isPal(n) ?>printf>(>'Yes
'>) :>printf>(>'No
'>);> > >n = 8999;> >isPal(n) ?>printf>(>'Yes
'>) :>printf>(>'No
'>);> > >return> 0;> }> |
>
>
Java
// A recursive Java program to> // check whether a given number> // is palindrome or not> import> java.io.*;> import> java.util.*;> > public> class> CheckPalindromeNumberRecursion {> > >// A function that returns true> >// only if num contains one digit> >public> static> int> oneDigit(>int> num) {> > >if> ((num>=>0>) && (num <>10>))> >return> 1>;> >else> >return> 0>;> >}> > >public> static> int> isPalUtil> >(>int> num,>int> dupNum)>throws> Exception {> > >// base condition to return once we> >// move past first digit> >if> (num ==>0>) {> >return> dupNum;> >}>else> {> >dupNum = isPalUtil(num />10>, dupNum);> >}> > >// Check for equality of first digit of> >// num and dupNum> >if> (num %>10> == dupNum %>10>) {> >// if first digit values of num and> >// dupNum are equal divide dupNum> >// value by 10 to keep moving in sync> >// with num.> >return> dupNum />10>;> >}>else> {> >// At position values are not> >// matching throw exception and exit.> >// no need to proceed further.> >throw> new> Exception();> >}> > >}> > >public> static> int> isPal(>int> num)> >throws> Exception {> > >if> (num <>0>)> >num = (-num);> > >int> dupNum = (num);> > >return> isPalUtil(num, dupNum);> >}> > >public> static> void> main(String args[]) {> > >int> n =>12421>;> >try> {> >isPal(n);> >System.out.println(>'Yes'>);> >}>catch> (Exception e) {> >System.out.println(>'No'>);> >}> >n =>1231>;> >try> {> >isPal(n);> >System.out.println(>'Yes'>);> >}>catch> (Exception e) {> >System.out.println(>'No'>);> >}> > >n =>12>;> >try> {> >isPal(n);> >System.out.println(>'Yes'>);> >}>catch> (Exception e) {> >System.out.println(>'No'>);> >}> > >n =>88>;> >try> {> >isPal(n);> >System.out.println(>'Yes'>);> >}>catch> (Exception e) {> >System.out.println(>'No'>);> >}> > >n =>8999>;> >try> {> >isPal(n);> >System.out.println(>'Yes'>);> >}>catch> (Exception e) {> >System.out.println(>'No'>);> >}> >}> }> > // This code is contributed> // by Nasir J> |
>
>
Python3
# A recursive Python3 program to check> # whether a given number is palindrome or not> > # A function that returns true> # only if num contains one digit> def> oneDigit(num):> > ># comparison operation is faster> ># than division operation. So> ># using following instead of> ># 'return num / 10 == 0;'> >return> ((num>>>)>and> >(num <>10>))> > # A recursive function to find> # out whether num is palindrome> # or not. Initially, dupNum> # contains address of a copy of num.> def> isPalUtil(num, dupNum):> > ># Base case (needed for recursion> ># termination): This statement> ># mainly compares the first digit> ># with the last digit> >if> oneDigit(num):> >return> (num>=>=> (dupNum[>0>])>%> 10>)> > ># This is the key line in this> ># method. Note that all recursive> ># calls have a separate copy of> ># num, but they all share same> ># copy of *dupNum. We divide num> ># while moving up the recursion tree> >if> not> isPalUtil(num>/>/>10>, dupNum):> >return> False> > ># The following statements are> ># executed when we move up the> ># recursion call tree> >dupNum[>0>]>=> dupNum[>0>]>/>/>10> > ># At this point, if num%10> ># contains i'th digit from> ># beginning, then (*dupNum)%10> ># contains i'th digit from end> >return> (num>%> 10> =>=> (dupNum[>0>])>%> 10>)> > # The main function that uses> # recursive function isPalUtil()> # to find out whether num is> # palindrome or not> def> isPal(num):> ># If num is negative,> ># make it positive> >if> (num <>0>):> >num>=> (>->num)> > ># Create a separate copy of> ># num, so that modifications> ># made to address dupNum> ># don't change the input number.> >dupNum>=> [num]># *dupNum = num> > >return> isPalUtil(num, dupNum)> > # Driver Code> n>=> 12321> if> isPal(n):> >print>(>'Yes'>)> else>:> >print>(>'No'>)> > n>=> 12> if> isPal(n) :> >print>(>'Yes'>)> else>:> >print>(>'No'>)> > n>=> 88> if> isPal(n) :> >print>(>'Yes'>)> else>:> >print>(>'No'>)> > n>=> 8999> if> isPal(n) :> >print>(>'Yes'>)> else>:> >print>(>'No'>)> > # This code is contributed by mits> |
>
dvejetainis medis inorder traversal
>
C#
// A recursive C# program to> // check whether a given number> // is palindrome or not> using> System;> > class> GFG> {> > // A function that returns true> // only if num contains one digit> public> static> int> oneDigit(>int> num)> {> >// comparison operation is> >// faster than division> >// operation. So using> >// following instead of> >// 'return num / 10 == 0;'> >if>((num>= 0) &&(skaičius<10))> >return> 1;> >else> >return> 0;> }> > // A recursive function to> // find out whether num is> // palindrome or not.> // Initially, dupNum contains> // address of a copy of num.> public> static> int> isPalUtil(>int> num,> >int> dupNum)> {> >// Base case (needed for recursion> >// termination): This statement> >// mainly compares the first digit> >// with the last digit> >if> (oneDigit(num) == 1)> >if>(num == (dupNum) % 10)> >return> 1;> >else> >return> 0;> > >// This is the key line in> >// this method. Note that> >// all recursive calls have> >// a separate copy of num,> >// but they all share same> >// copy of *dupNum. We divide> >// num while moving up the> >// recursion tree> >if> (isPalUtil((>int>)(num / 10), dupNum) == 0)> >return> -1;> > >// The following statements> >// are executed when we move> >// up the recursion call tree> >dupNum = (>int>)(dupNum / 10);> > >// At this point, if num%10> >// contains i'th digit from> >// beginning, then (*dupNum)%10> >// contains i'th digit from end> >if>(num % 10 == (dupNum) % 10)> >return> 1;> >else> >return> 0;> }> > // The main function that uses> // recursive function isPalUtil()> // to find out whether num is> // palindrome or not> public> static> int> isPal(>int> num)> {> >// If num is negative,> >// make it positive> >if> (num <0)> >num = (-num);> > >// Create a separate copy> >// of num, so that modifications> >// made to address dupNum> >// don't change the input number.> >int> dupNum = (num);>// *dupNum = num> > >return> isPalUtil(num, dupNum);> }> > // Driver Code> public> static> void> Main()> {> int> n = 12321;> if>(isPal(n) == 0)> >Console.WriteLine(>'Yes'>);> else> >Console.WriteLine(>'No'>);> > n = 12;> if>(isPal(n) == 0)> >Console.WriteLine(>'Yes'>);> else> >Console.WriteLine(>'No'>);> > n = 88;> if>(isPal(n) == 1)> >Console.WriteLine(>'Yes'>);> else> >Console.WriteLine(>'No'>);> > n = 8999;> if>(isPal(n) == 0)> >Console.WriteLine(>'Yes'>);> else> >Console.WriteLine(>'No'>);> }> }> > // This code is contributed by mits> |
>
>
Javascript
> // A recursive javascript program to> // check whether a given number> // is palindrome or not> > >// A function that returns true> >// only if num contains one digit> >function> oneDigit(num) {> > >if> ((num>= 0) && (skaičius<10))> >return> 1;> >else> >return> 0;> >}> > >function> isPalUtil> >(num , dupNum) {> > >// base condition to return once we> >// move past first digit> >if> (num == 0) {> >return> dupNum;> >}>else> {> >dupNum = isPalUtil(parseInt(num / 10), dupNum);> >}> > >// Check for equality of first digit of> >// num and dupNum> >if> (num % 10 == dupNum % 10) {> >// if first digit values of num and> >// dupNum are equal divide dupNum> >// value by 10 to keep moving in sync> >// with num.> >return> parseInt(dupNum / 10);> >}>else> {> >// At position values are not> >// matching throw exception and exit.> >// no need to proceed further.> >throw> e;> >}> > >}> > >function> isPal(num)> >{> > >if> (num <0)> >num = (-num);> > >var> dupNum = (num);> > >return> isPalUtil(num, dupNum);> >}> > > > >var> n = 1242;> >try> {> >isPal(n);> >document.write(>' Yes'>);> >}>catch> (e) {> >document.write(>' No'>);> >}> >n = 1231;> >try> {> >isPal(n);> >document.write(>' Yes'>);> >}>catch> (e) {> >document.write(>' No'>);> >}> > >n = 12;> >try> {> >isPal(n);> >document.write(>' Yes'>);> >}>catch> (e) {> >document.write(>' No'>);> >}> > >n = 88;> >try> {> >isPal(n);> >document.write(>' Yes'>);> >}>catch> (e) {> >document.write(>' No'>);> >}> > >n = 8999;> >try> {> >isPal(n);> >document.write(>' Yes'>);> >}>catch> (e) {> >document.write(>' No'>);> >}> > // This code is contributed by Amit Katiyar> > |
>
>
PHP
// A recursive PHP program to // check whether a given number // is palindrome or not // A function that returns true // only if num contains one digit function oneDigit($num) { // comparison operation is faster // than division operation. So // using following instead of // 'return num / 10 == 0;' return (($num>= 0) && ($num<10)); } // A recursive function to find // out whether num is palindrome // or not. Initially, dupNum // contains address of a copy of num. function isPalUtil($num, $dupNum) { // Base case (needed for recursion // termination): This statement // mainly compares the first digit // with the last digit if (oneDigit($num)) return ($num == ($dupNum) % 10); // This is the key line in this // method. Note that all recursive // calls have a separate copy of // num, but they all share same // copy of *dupNum. We divide num // while moving up the recursion tree if (!isPalUtil((int)($num / 10), $dupNum)) return -1; // The following statements are // executed when we move up the // recursion call tree $dupNum = (int)($dupNum / 10); // At this point, if num%10 // contains i'th digit from // beginning, then (*dupNum)%10 // contains i'th digit from end return ($num % 10 == ($dupNum) % 10); } // The main function that uses // recursive function isPalUtil() // to find out whether num is // palindrome or not function isPal($num) { // If num is negative, // make it positive if ($num <0) $num = (-$num); // Create a separate copy of // num, so that modifications // made to address dupNum // don't change the input number. $dupNum = ($num); // *dupNum = num return isPalUtil($num, $dupNum); } // Driver Code $n = 12321; if(isPal($n) == 0) echo 'Yes
'; else echo 'No
'; $n = 12; if(isPal($n) == 0) echo 'Yes
'; else echo 'No
'; $n = 88; if(isPal($n) == 1) echo 'Yes
'; else echo 'No
'; $n = 8999; if(isPal($n) == 0) echo 'Yes
'; else echo 'No
'; // This code is contributed by m_kit ?>>> |
>Yes No Yes No>
Laiko sudėtingumas: O(log n)
Pagalbinė erdvė: O(log n)
Norėdami patikrinti, ar numeris yra palindromas, ar ne, nenaudojant papildomos vietos
2 metodas: String() metodo naudojimas
- Kai to skaičiaus skaitmenų skaičius viršija 1018, negalime priimti to skaičiaus kaip sveikojo skaičiaus, nes ilgio ilgio int diapazonas neatitinka nurodyto skaičiaus.
- Taigi imkite įvestį kaip eilutę, paleiskite ciklą nuo pradžios iki ilgio/2 ir patikrinkite pirmąjį simbolį (skaitmenį) iki paskutinio eilutės simbolio ir nuo antrojo iki antrojo paskutinio ir t. t. .... Jei simboliai nesutampa, eilutė nebūtų palindromas.
Žemiau pateikiamas pirmiau minėto metodo įgyvendinimas
C++14
// C++ implementation of the above approach> #include> using> namespace> std;> > // Function to check palindrome> int> checkPalindrome(string str)> {> >// Calculating string length> >int> len = str.length();> > >// Traversing through the string> >// upto half its length> >for> (>int> i = 0; i // Comparing i th character // from starting and len-i // th character from end if (str[i] != str[len - i - 1]) return false; } // If the above loop doesn't return then it is // palindrome return true; } // Driver Code int main() { // taking number as string string st = '112233445566778899000000998877665544332211'; if (checkPalindrome(st) == true) cout << 'Yes'; else cout << 'No'; return 0; } // this code is written by vikkycirus> |
>
>
Java
// Java implementation of the above approach> import> java.io.*;> > class> GFG{> > // Function to check palindrome> static> boolean> checkPalindrome(String str)> {> > >// Calculating string length> >int> len = str.length();> > >// Traversing through the string> >// upto half its length> >for>(>int> i =>0>; i 2; i++) { // Comparing i th character // from starting and len-i // th character from end if (str.charAt(i) != str.charAt(len - i - 1)) return false; } // If the above loop doesn't return then // it is palindrome return true; } // Driver Code public static void main(String[] args) { // Taking number as string String st = '112233445566778899000000998877665544332211'; if (checkPalindrome(st) == true) System.out.print('Yes'); else System.out.print('No'); } } // This code is contributed by subhammahato348> |
>
>
Python3
# Python3 implementation of the above approach> > # function to check palindrome> def> checkPalindrome(>str>):> > ># Run loop from 0 to len/2> >for> i>in> range>(>0>,>len>(>str>)>/>/>2>):> >if> str>[i] !>=> str>[>len>(>str>)>->i>->1>]:> >return> False> > ># If the above loop doesn't> >#return then it is palindrome> >return> True> > > # Driver code> st>=> '112233445566778899000000998877665544332211'> if>(checkPalindrome(st)>=>=> True>):> >print>(>'it is a palindrome'>)> else>:> >print>(>'It is not a palindrome'>)> |
>
>
C#
// C# implementation of the above approach> using> System;> > class> GFG{> > // Function to check palindrome> static> bool> checkPalindrome(>string> str)> {> > >// Calculating string length> >int> len = str.Length;> > >// Traversing through the string> >// upto half its length> >for>(>int> i = 0; i { // Comparing i th character // from starting and len-i // th character from end if (str[i] != str[len - i - 1]) return false; } // If the above loop doesn't return then // it is palindrome return true; } // Driver Code public static void Main() { // Taking number as string string st = '112233445566778899000000998877665544332211'; if (checkPalindrome(st) == true) Console.Write('Yes'); else Console.Write('No'); } } // This code is contributed by subhammahato348> |
>
>
Javascript
> > // Javascript implementation of the above approach> > // Function to check palindrome> function> checkPalindrome(str)> {> >// Calculating string length> >var> len = str.length;> > >// Traversing through the string> >// upto half its length> >for> (>var> i = 0; i // Comparing ith character // from starting and len-ith // character from end if (str[i] != str[len - i - 1]) return false; } // If the above loop doesn't return then it is // palindrome return true; } // Driver Code // taking number as string let st = '112233445566778899000000998877665544332211'; if (checkPalindrome(st) == true) document.write('Yes'); else document.write('No'); // This code is contributed by Mayank Tyagi> |
>
>Išvestis
Yes>
Laiko sudėtingumas: O(|str|)
Pagalbinė erdvė : O(1)
3 būdas:
Čia yra paprasčiausias būdas patikrinti, ar skaičius yra palindromas, ar ne. Šis metodas gali būti naudojamas, kai skaitmenų skaičius pateiktame skaičiuje yra mažesnis nei 10^18, nes jei to skaičiaus skaitmenų skaičius viršija 10^18, to skaičiaus negalime laikyti sveikojo skaičiaus, nes ilgio ilgio diapazonas int neatitinka nurodyto skaičiaus.
Norėdami patikrinti, ar pateiktas skaičius yra palindromas, ar ne, mes tiesiog apverssime duoto skaičiaus skaitmenis ir patikrinsime, ar to skaičiaus atvirkštinė dalis yra lygi pradiniam skaičiui, ar ne. Jei atvirkštinis skaičius yra lygus tam skaičiui, skaičius bus palindromas, kitaip jis nebus palindromas.
C++
// C++ program to check if a number is Palindrome> #include> using> namespace> std;> // Function to check Palindrome> bool> checkPalindrome(>int> n)> {> >int> reverse = 0;> >int> temp = n;> >while> (temp != 0) {> >reverse = (reverse * 10) + (temp % 10);> >temp = temp / 10;> >}> >return> (reverse> >== n);>// if it is true then it will return 1;> >// else if false it will return 0;> }> int> main()> {> >int> n = 7007;> >if> (checkPalindrome(n) == 1) {> >cout <<>'Yes
'>;> >}> >else> {> >cout <<>'No
'>;> >}> >return> 0;> }> // This code is contributed by Suruchi Kumari> |
>
>
Java
/*package whatever //do not write package name here */> > import> java.io.*;> > class> GFG {> >// Java program to check if a number is Palindrome> > >// Function to check Palindrome> >static> boolean> checkPalindrome(>int> n)> >{> >int> reverse =>0>;> >int> temp = n;> >while> (temp !=>0>) {> >reverse = (reverse *>10>) + (temp %>10>);> >temp = temp />10>;> >}> >return> (reverse == n);>// if it is true then it will return 1;> >// else if false it will return 0;> >}> > >// Driver Code> >public> static> void> main(String args[])> >{> >int> n =>7007>;> >if> (checkPalindrome(n) ==>true>) {> >System.out.println(>'Yes'>);> >}> >else> {> >System.out.println(>'No'>);> >}> >}> }> > // This code is contributed by shinjanpatra> |
>
>
css wrap tekstas
Python3
# Python3 program to check if a number is Palindrome> > # Function to check Palindrome> def> checkPalindrome(n):> > >reverse>=> 0> >temp>=> n> >while> (temp !>=> 0>):> >reverse>=> (reverse>*> 10>)>+> (temp>%> 10>)> >temp>=> temp>/>/> 10> > >return> (reverse>=>=> n)># if it is true then it will return 1;> ># else if false it will return 0;> > # driver code> n>=> 7007> if> (checkPalindrome(n)>=>=> 1>):> >print>(>'Yes'>)> > else>:> >print>(>'No'>)> > # This code is contributed by shinjanpatra> |
>
>
C#
// C# program to check if a number is Palindrome> > using> System;> > class> GFG {> > >// Function to check Palindrome> >static> bool> checkPalindrome(>int> n)> >{> >int> reverse = 0;> >int> temp = n;> >while> (temp != 0) {> >reverse = (reverse * 10) + (temp % 10);> >temp = temp / 10;> >}> >return> (> >reverse> >== n);>// if it is true then it will return 1;> >// else if false it will return 0;> >}> > >// Driver Code> >public> static> void> Main(>string>[] args)> >{> >int> n = 7007;> >if> (checkPalindrome(n) ==>true>) {> >Console.WriteLine(>'Yes'>);> >}> >else> {> >Console.WriteLine(>'No'>);> >}> >}> }> > // This code is contributed by phasing17> |
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Javascript
> > // JavaScript program to check if a number is Palindrome> > // Function to check Palindrome> function> checkPalindrome(n)> {> >let reverse = 0;> >let temp = n;> >while> (temp != 0) {> >reverse = (reverse * 10) + (temp % 10);> >temp = Math.floor(temp / 10);> >}> >return> (reverse == n);>// if it is true then it will return 1;> >// else if false it will return 0;> }> > // driver code> > let n = 7007;> if> (checkPalindrome(n) == 1) {> >document.write(>'Yes'>,>''>);> }> else> {> >document.write(>'No'>,>''>);> }> > > // This code is contributed by shinjanpatra> > > |
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>Išvestis
Yes>
Laiko sudėtingumas: O(log10(n)) arba O (skaitmenų skaičius tam tikrame skaičiuje)
Pagalbinė erdvė : O(1) arba konstanta
Šį straipsnį parengėAshish Barnwal.