Dada una string, escriba una función recursiva que verifique si la string dada es un palíndromo, de lo contrario, no un palíndromo.
Ejemplos:
Input : malayalam Output : Yes Reverse of malayalam is also malayalam. Input : max Output : No Reverse of max is not max.
Hemos discutido una función iterativa aquí .
La idea de una función recursiva es simple:
1) If there is only one character in string return true. 2) Else compare first and last characters and recur for remaining substring.
A continuación se muestra la implementación de la idea anterior:
C++
// A recursive C++ program to
// check whether a given number
// is palindrome or not
#include <bits/stdc++.h>
using namespace std;
// A recursive function that
// check a str[s..e] is
// palindrome or not.
bool isPalRec(char str[],
int s, int e)
{
// If there is only one character
if (s == e)
return true;
// If first and last
// characters do not match
if (str[s] != str[e])
return false;
// If there are more than
// two characters, check if
// middle substring is also
// palindrome or not.
if (s < e + 1)
return isPalRec(str, s + 1, e - 1);
return true;
}
bool isPalindrome(char str[])
{
int n = strlen(str);
// An empty string is
// considered as palindrome
if (n == 0)
return true;
return isPalRec(str, 0, n - 1);
}
// Driver Code
int main()
{
char str[] = "geeg";
if (isPalindrome(str))
cout << "Yes";
else
cout << "No";
return 0;
}
// This code is contributed by shivanisinghss2110
C
// A recursive C program to
// check whether a given number
// is palindrome or not
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
// A recursive function that
// check a str[s..e] is
// palindrome or not.
bool isPalRec(char str[],
int s, int e)
{
// If there is only one character
if (s == e)
return true;
// If first and last
// characters do not match
if (str[s] != str[e])
return false;
// If there are more than
// two characters, check if
// middle substring is also
// palindrome or not.
if (s < e + 1)
return isPalRec(str, s + 1, e - 1);
return true;
}
bool isPalindrome(char str[])
{
int n = strlen(str);
// An empty string is
// considered as palindrome
if (n == 0)
return true;
return isPalRec(str, 0, n - 1);
}
// Driver Code
int main()
{
char str[] = "geeg";
if (isPalindrome(str))
printf("Yes");
else
printf("No");
return 0;
}
Java
// A recursive JAVA program to
// check whether a given String
// is palindrome or not
import java.io.*;
class GFG
{
// A recursive function that
// check a str(s..e) is
// palindrome or not.
static boolean isPalRec(String str,
int s, int e)
{
// If there is only one character
if (s == e)
return true;
// If first and last
// characters do not match
if ((str.charAt(s)) != (str.charAt(e)))
return false;
// If there are more than
// two characters, check if
// middle substring is also
// palindrome or not.
if (s < e + 1)
return isPalRec(str, s + 1, e - 1);
return true;
}
static boolean isPalindrome(String str)
{
int n = str.length();
// An empty string is
// considered as palindrome
if (n == 0)
return true;
return isPalRec(str, 0, n - 1);
}
// Driver Code
public static void main(String args[])
{
String str = "geeg";
if (isPalindrome(str))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed
// by Nikita Tiwari
Python
# A recursive Python program # to check whether a given # number is palindrome or not # A recursive function that # check a str[s..e] is # palindrome or not. def isPalRec(st, s, e) : # If there is only one character if (s == e): return True # If first and last # characters do not match if (st[s] != st[e]) : return False # If there are more than # two characters, check if # middle substring is also # palindrome or not. if (s < e + 1) : return isPalRec(st, s + 1, e - 1); return True def isPalindrome(st) : n = len(st) # An empty string is # considered as palindrome if (n == 0) : return True return isPalRec(st, 0, n - 1); # Driver Code st = "geeg" if (isPalindrome(st)) : print "Yes" else : print "No" # This code is contributed # by Nikita Tiwari.
C#
// A recursive C# program to
// check whether a given number
// is palindrome or not
using System;
class GFG
{
// A recursive function that
// check a str(s..e)
// is palindrome or not.
static bool isPalRec(String str,
int s,
int e)
{
// If there is only one character
if (s == e)
return true;
// If first and last character
// do not match
if ((str[s]) != (str[e]))
return false;
// If there are more than two
// characters, check if middle
// substring is also
// palindrome or not.
if (s < e + 1)
return isPalRec(str, s + 1,
e - 1);
return true;
}
static bool isPalindrome(String str)
{
int n = str.Length;
// An empty string is considered
// as palindrome
if (n == 0)
return true;
return isPalRec(str, 0, n - 1);
}
// Driver Code
public static void Main()
{
String str = "geeg";
if (isPalindrome(str))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by Nitin Mittal.
PHP
<?php
// A recursive php program to
// check whether a given number
// is palindrome or not
// A recursive function that
// check a str[s..e] is
// palindrome or not.
function isPalRec($str, $s,$e)
{
// If there is only one character
if ($s == $e)
return true;
// If first and last
// characters do not match
if ($str[$s] != $str[$e])
return false;
// If there are more than two
// characters, check if middle
// substring is also palindrome or not.
if ($s < $e + 1)
return isPalRec($str, $s + 1, $e - 1);
return true;
}
function isPalindrome($str)
{
$n = strlen($str);
// An empty string is
// considered as palindrome
if ($n == 0)
return true;
return isPalRec($str, 0, $n - 1);
}
// Driver Code
{
$str = "geeg";
if (isPalindrome($str))
echo("Yes");
else
echo("No");
return 0;
}
// This code is contributed
// by nitin mittal.
?>
Javascript
<script>
// A recursive javascript program to
// check whether a given String
// is palindrome or not
// A recursive function that
// check a str(s..e) is
// palindrome or not.
function isPalRec( str , s , e) {
// If there is only one character
if (s == e)
return true;
// If first and last
// characters do not match
if ((str.charAt(s)) != (str.charAt(e)))
return false;
// If there are more than
// two characters, check if
// middle substring is also
// palindrome or not.
if (s < e + 1)
return isPalRec(str, s + 1, e - 1);
return true;
}
function isPalindrome( str) {
var n = str.length;
// An empty string is
// considered as palindrome
if (n == 0)
return true;
return isPalRec(str, 0, n - 1);
}
// Driver Code
var str = "geeg";
if (isPalindrome(str))
document.write("Yes");
else
document.write("No");
// This code contributed by gauravrajput1
</script>
Producción
Yes
Otro enfoque :
Básicamente, mientras atraviesa, verifique si el índice i-th y ni-1th son iguales o no.
Si no son iguales devuelve falso y si son iguales entonces continúa con las llamadas de recurrencia.
C++
#include <iostream>
using namespace std;
bool isPalindrome(string s, int i){
if(i > s.size()/2){
return true ;
}
return s[i] == s[s.size()-i-1] && isPalindrome(s, i+1) ;
}
int main()
{
string str = "geeg" ;
if (isPalindrome(str, 0))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
/*package whatever //do not write package name here */
import java.io.*;
class GFG {
public static boolean isPalindrome(String s, int i){
if(i > s.length()/2)
{
return true ;
}
return s.charAt(i) == s.charAt(s.length()-i-1) && isPalindrome(s, i+1) ;
}
public static void main (String[] args) {
String str = "geeg" ;
if (isPalindrome(str, 0))
{ System.out.println("Yes"); }
else
{ System.out.println("No"); }
}
}
// This code is contributed by akashish.
C#
using System;
public class GFG{
public static bool isPalindrome(string s, int i){
if(i > s.Length/2){
return true ;
}
return s[i] == s[s.Length-i-1] && isPalindrome(s, i+1) ;
}
public static void Main (){
// Code
string str = "geeg" ;
if (isPalindrome(str, 0))
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
// This code is contributed by akashish_.
Javascript
<script>
function isPalindrome(s,i){
if(i > s.length/2)
{return true;}
return s[i] == s[s.length-i-1] && isPalindrome(s, i+1)
}
let str = "geeg";
let ans = isPalindrome(str, 0);
if (ans == true)
{
console.log("Yes");}
else
{
console.log("No");}
// This code is contributed by akashish__
</script>
Producción
Yes
Complejidad temporal: O(n)
Espacio auxiliar: O(n)
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA