Ak je dané kladné celé číslo, napíšte funkciu, ktorá vráti hodnotu true, ak je zadané číslo palindróm, inak hodnotu false. Napríklad 12321 je palindróm, ale 1451 nie je palindróm.
Metóda 1:
Nech je dané číslo na jeden . Jednoduchý spôsob riešenia tohto problému je prvý obrátené číslice na jeden , potom porovnajte opak na jeden s na jeden . Ak sú obe rovnaké, vráti hodnotu true, inak false.
Nasleduje zaujímavá metóda inšpirovaná metódou č. 2 toto príspevok. Cieľom je vytvoriť kópiu na jeden a rekurzívne odovzdať kópiu odkazom a prejsť na jeden podľa hodnoty. V rekurzívnych hovoroch rozdeľte na jeden o 10 pri pohybe nadol v strome rekurzie. Pri pohybe nahor v strome rekurzie vydeľte kópiu 10. Keď sa stretnú vo funkcii, pre ktorú sú ukončené všetky podriadené volania, posledná číslica na jeden bude itá číslica od začiatku a posledná číslica kópie bude itá číslica od konca.
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 && č<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 && č<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>>=> 0>)>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> |
>
>
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) &&(č<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> |
>
java bublinové triedenie
>
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) && (č<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 ?>> |
>
>Výkon
Yes No Yes No>
Časová zložitosť: O (log n)
Pomocný priestor: O (log n)
Ak chcete skontrolovať, či je číslo palindróm alebo nie, bez použitia medzery navyše
Metóda 2: Použitie metódy string().
abstraktné metódy
- Keď počet číslic tohto čísla presiahne 1018, nemôžeme toto číslo považovať za celé číslo, pretože rozsah long long int nevyhovuje danému číslu.
- Takže vezmite vstup ako reťazec, Spustite cyklus od začiatku po dĺžku/2 a skontrolujte prvý znak (číselný) po posledný znak reťazca a predposledný, atď.... Ak sa nejaký znak nezhoduje, reťazec nebol by to palindróm.
Nižšie je uvedená implementácia vyššie uvedeného prístupu
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> |
>
>Výkon
Yes>
Časová zložitosť: O(|str|)
Pomocný priestor : O(1)
Metóda 3:
Tu je najjednoduchší spôsob, ako skontrolovať, či je číslo Palindróm alebo nie. Tento prístup je možné použiť, keď je počet číslic v danom čísle menší ako 10^18, pretože ak počet číslic tohto čísla presiahne 10^18, nemôžeme toto číslo považovať za celé číslo, pretože rozsah long long int nevyhovuje danému číslu.
Aby sme skontrolovali, či dané číslo je palindróm alebo nie, jednoducho prehodíme číslice daného čísla a skontrolujeme, či sa rub tohto čísla rovná pôvodnému číslu alebo nie. Ak sa opak čísla rovná tomuto číslu, číslo bude palindróm, inak to nebude palindróm.
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> |
mysql vytvoriť používateľa
>
>
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> |
>
>
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> > > |
>
>Výkon
Yes>
Časová zložitosť: O(log10(n)) alebo O (počet číslic v danom čísle)
Pomocný priestor : O(1) alebo konštanta
Tento článok zostavujeAshish Barnwal.