cls
input "enter a string=";s$
l=len(s$)
for i = l to 1 step-1
c$=mid$(s$,i,l)
next i
r$=c$+r$
if r$=c$ then
print "no. is palindrome"
else
print "no. is not palindrome"
endif
end
/*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 (); }
Palindrome number is a number like 121 which remains the same when its digits are reversed. To find this number in a simple java program, just follow the below way. sum = 0; while(n>0) { r=n % 10; sum=concat(r); n=n / 10; } print r;
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"); }}
This program only suits PHP. If you want a proper one try C program for it available on web <body> <?php if(isset($_POST['submit'])) { $text = $_POST['text']; $string = mysql_real_escape_string($text); $invert = strrev($string); if($string == $invert) { echo "<br>Given number is a palindrome!!"; } else { echo "<br>Given number is not a palindrome!!"; } } ?> <form method="post"> <input type="text" name="text" id="text" /> <input type="submit" name="submit" value="Submit" id="submit" onclick="<?php $_SERVER['PHP_SELF']; ?>" /> </form> </body>
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You can do this: <?php if ( $word === strrev( $word ) ) { echo "The word is a palindrome"; } else { echo "The word is not a palindrome"; }
Write your program and if you are having a problem post it here with a description of the problem you are having. What you are asking is for someone to do your homework for you.
It is a simple program. i think u may understand it :#include#include#includevoid main(){char s[10]=answers.com;char x[10];int a;clrscr();strcpy(x,s);strrev(s);a=strcmp(s,x);if(a==0){printf("the entered string is palindrome");}else{printf("the entered string is not palindrome");}output:given string is not palindrome
/*To check whether a string is palindrome*/includeincludevoid main () { int i,j,f=0; char a[10]; clrscr (); gets(a); for (i=0;a[i]!='\0';i++) { } i--; for (j=0;a[j]!='\0';j++,i--) { if (a[i]!=a[j]) f=1; } if (f==0) printf("string is palindrome"); else printf("string is not palindrome"); getch (); }
To check if a string is a palindrome, point to each end of the string and work inwards towards the middle. If the characters pointed at differ, the string is not a palindrome. When the pointers meet or cross each other, the string is a palindrome. Note that the string cannot contain whitespace or punctuation and comparisons must not be case-sensitive.
You could use a function like this:function isPalindrome($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 $test.' is not a palindrome';}
Palindrome number is a number like 121 which remains the same when its digits are reversed. To find this number in a simple java program, just follow the below way. sum = 0; while(n>0) { r=n % 10; sum=concat(r); n=n / 10; } print r;
import java.util.Scanner; public class Palindrome{ public static void main(String[] args){ String front; String back =""; char[] failure; String backwards; Scanner input=new Scanner(System.in); System.out.print("Enter a word: "); front=input.next(); front=front.replaceAll(" ", ""); failure=front.toCharArray(); for (int i=0; i<failure.length; i++){ back=failure[i] + back; } if (front.equals(back)){ System.out.print("That word is a palindrome"); }else System.out.print("That word is not a palindrome"); }}
Copy and reverse the string. If the reversed string is equal to the original string, the string is a palindrome, otherwise it is not. When working with strings that hold natural language phrases (including punctuation, whitespace and so on) we must remove all the non-alphanumerics and convert the remainder to a common case, such as lower-case, prior to copying and reversing the string.
This program only suits PHP. If you want a proper one try C program for it available on web <body> <?php if(isset($_POST['submit'])) { $text = $_POST['text']; $string = mysql_real_escape_string($text); $invert = strrev($string); if($string == $invert) { echo "<br>Given number is a palindrome!!"; } else { echo "<br>Given number is not a palindrome!!"; } } ?> <form method="post"> <input type="text" name="text" id="text" /> <input type="submit" name="submit" value="Submit" id="submit" onclick="<?php $_SERVER['PHP_SELF']; ?>" /> </form> </body>
If you want to check whether a string is a palindrome, you can reverse the string (for example, the Java class StringBuffer has a reverse() method), and then compare whether the two strings - the original string and the reverted string - are equal. Alternately, you could write a loop that checks whether the first character of the string is equal to the last one, the second is equal to the second-last one, etc.; that is, you have a counter variable (in a "for" loop) that goes from zero to length - 1 (call it "i"), and compare character #i with character #(length-i-1) inside the loop.
Use the Exception class and type converters.