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.
def isPalindrome(s): return s == s[::-1] then just call the function with a string like isPalindrome('poop')
sdfdg
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;
You don't have to write palindrome programs in wikipedia.
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>
You can do this: <?php if ( $word === strrev( $word ) ) { echo "The word is a palindrome"; } else { echo "The 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
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.
def isPalindrome(s): return s == s[::-1] then just call the function with a string like isPalindrome('poop')
sdfdg
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.
#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(); }
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;
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.
You don't have to write palindrome programs in wikipedia.
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>
#include <stdio.h> #include <string.h> #define N 100 #define PALINDROME 0 #define NONPALINDROME 1 /*Program that tells you whether what you enter is a palindrome or not*/ char pal[N]; //input line int i; //counter int k; //counter int tag; int flag = PALINDROME; /*flag=0 ->palindrome, flag =1 ->nonpalindrome*/ int main() { printf("Enter something: \n"); scanf("%s", pal); tag = strlen(pal); /*determine length of string*/ /* pointer running from the beginning and the end simultaneously looking for two characters that don't match */ /* initially assumed that string IS a palindrome */ /* the following for loop looks for inequality and flags the string as a non palindrome the moment two characters don't match */ for (i=0,k=tag-1; i=0; i++,k--) { if(pal[i] != pal[k]) { flag=NONPALINDROME; break; } } if(flag == PALINDROME) { printf("This is a palindrome\n"); } else { printf("This is NOT a palindrome\n"); } return 0; } #include <stdio.h> #include <string.h> #define N 100 #define PALINDROME 0 #define NONPALINDROME 1 /*Program that tells you whether what you enter is a palindrome or not*/ char pal[N]; //input line int i; //counter int k; //counter int tag; int flag = PALINDROME; /*flag=0 ->palindrome, flag =1 ->nonpalindrome*/ int main() { printf("Enter something: \n"); scanf("%s", pal); tag = strlen(pal); /*determine length of string*/ /* pointer running from the beginning and the end simultaneously looking for two characters that don't match */ /* initially assumed that string IS a palindrome */ /* the following for loop looks for inequality and flags the string as a non palindrome the moment two characters don't match */ for (i=0,k=tag-1; i=0; i++,k--) { if(pal[i] != pal[k]) { flag=NONPALINDROME; break; } } if(flag == PALINDROME) { printf("This is a palindrome\n"); } else { printf("This is NOT a palindrome\n"); } return 0; }