len=0
i=1
echo -n "Enter a String: "
read str
len=`echo $str | wc -c`
len=`expr $len - 1`
halfLen=`expr $len / 2`
while [ $i -le $halfLen ]
do
c1=`echo $str|cut -c$i`
c2=`echo $str|cut -c$len`
if [ $c1 != $c2 ] ; then
echo "string is not palindrome"
exit
fi
i=`expr $i + 1`
len=`expr $len - 1`
done
echo "String is Palindrome"
/*To check whether a string is palindrome*/includeincludevoid main () { int i,j,f=0; char a[10]; clrscr (); gets(a); for (i=0;a[i]!='\0';i++) { } i--; for (j=0;a[j]!='\0';j++,i--) { if (a[i]!=a[j]) f=1; } if (f==0) printf("string is palindrome"); else printf("string is not palindrome"); getch (); }
import java.util.Scanner; public class Palindrome{ public static void main(String[] args){ String front; String back =""; char[] failure; String backwards; Scanner input=new Scanner(System.in); System.out.print("Enter a word: "); front=input.next(); front=front.replaceAll(" ", ""); failure=front.toCharArray(); for (int i=0; i<failure.length; i++){ back=failure[i] + back; } if (front.equals(back)){ System.out.print("That word is a palindrome"); }else System.out.print("That word is not a palindrome"); }}
/*To check whether a string is palindrome*/includeincludevoid main () {int i,j,f=0;char a[10];clrscr ();gets(a);for (i=0;a[i]!='\0';i++){}i--;for (j=0;a[j]!='\0';j++,i--){if (a[i]!=a[j])f=1;}if (f==0)printf("string is palindrome");else printf("string is not palindrome");getch ();}
Private Sub Command1_Click() a = Text1.Text b = StrReverse(a) If a = b Then Print "palindrome" Else Print "not palindrome" End If End Sub
Prepare the string for processing: Remove all punctuation from the string (e.g., commas, hyphens, whitespace, etc). Convert to the same case (e.g., lower-case). Instantiate two pointers, one pointing at the first character, the other pointing at the last character. Process: If the two pointers are pointing at the same position or have crossed each other, the string is a palindrome. Otherwise, compare the characters being pointed at. If they are not equal, the string is not a palindrome. Otherwise, move both pointers one position towards the middle of the string and repeat the process.
len=0 i=1 echo -n "Enter a String: " read str len=`echo $str | wc -c` len=`expr $len - 1` halfLen=`expr $len / 2` while [ $i -le $halfLen ] do c1=`echo $str|cut -c$i` c2=`echo $str|cut -c$len` if [ $c1 != $c2 ] ; then echo "string is not palindrome" exit fi i=`expr $i + 1` len=`expr $len - 1` done echo "String is Palindrome"
/*To check whether a string is palindrome*/includeincludevoid main () { int i,j,f=0; char a[10]; clrscr (); gets(a); for (i=0;a[i]!='\0';i++) { } i--; for (j=0;a[j]!='\0';j++,i--) { if (a[i]!=a[j]) f=1; } if (f==0) printf("string is palindrome"); else printf("string is not palindrome"); getch (); }
To check if a string is a palindrome, point to each end of the string and work inwards towards the middle. If the characters pointed at differ, the string is not a palindrome. When the pointers meet or cross each other, the string is a palindrome. Note that the string cannot contain whitespace or punctuation and comparisons must not be case-sensitive.
You can do this: <?php if ( $word === strrev( $word ) ) { echo "The word is a palindrome"; } else { echo "The word is not a palindrome"; }
You could use a function like this:function isPalindrome($string) {$string = strtolower($string);return (strrev($string) == $string) ? true : false;}and then to check a palindrome call an if statement like so:if(isPalindrome($test)) {echo $test.' is a palindrome';}else {echo $test.' is not a palindrome';}
import java.util.Scanner; public class Palindrome{ public static void main(String[] args){ String front; String back =""; char[] failure; String backwards; Scanner input=new Scanner(System.in); System.out.print("Enter a word: "); front=input.next(); front=front.replaceAll(" ", ""); failure=front.toCharArray(); for (int i=0; i<failure.length; i++){ back=failure[i] + back; } if (front.equals(back)){ System.out.print("That word is a palindrome"); }else System.out.print("That word is not a palindrome"); }}
It is a simple program. i think u may understand it :#include#include#includevoid main(){char s[10]=answers.com;char x[10];int a;clrscr();strcpy(x,s);strrev(s);a=strcmp(s,x);if(a==0){printf("the entered string is palindrome");}else{printf("the entered string is not palindrome");}output:given string is not palindrome
/*To check whether a string is palindrome*/includeincludevoid main () {int i,j,f=0;char a[10];clrscr ();gets(a);for (i=0;a[i]!='\0';i++){}i--;for (j=0;a[j]!='\0';j++,i--){if (a[i]!=a[j])f=1;}if (f==0)printf("string is palindrome");else printf("string is not palindrome");getch ();}
Copy and reverse the string. If the reversed string is equal to the original string, the string is a palindrome, otherwise it is not. When working with strings that hold natural language phrases (including punctuation, whitespace and so on) we must remove all the non-alphanumerics and convert the remainder to a common case, such as lower-case, prior to copying and reversing the string.
Private Sub Command1_Click() a = Text1.Text b = StrReverse(a) If a = b Then Print "palindrome" Else Print "not palindrome" End If End Sub
Check below link
#include <stdio.h> #include <conio.h> #include <string.h> void input(char a[ ]) { int i; printf("\n enter string\n"); scanf("%s",a); } void output(char a[ ]) { printf("\n string is %s",a); } int palindrome(char a[ ]) { int n,i; n=count(a); n=n-1; i=0; for(;a[n]==a[i] && n>=i;i++,n--); if(n>=i) return 0; else return 1; } void main( ) { char a[80],b[80],s; int n; printf("\n check palindrome"); input(a); n=palindrome(a); output(a); if(n==1) printf("\n palindrome"); else printf("\n not palindrome"); getch(); }