CareerCup

1. get to middle element of string.
2. take two pointers p1 and p2 point to middle element(if string is of odd length, or two different middle elements incase of even length).
3. while (p1 >= start, p2 <= end)
{
if (*p1 !=*p2)
return FALSE;
p1--;
p2++;
}
return true;

Code:
int len = strlen( str );
for(int i = 0; i< len; i++)
{
if(a[i] != a[len-1-i])
flag = 1;
}

if( flag )
Not a palindrome
else
A palindrome.

Complexity:
O(n)

package com.palindrometest;

public class PalindromeDate {

public static void main(String[] args) {

String input = "ISUDUSR";
boolean isPalindrome = false;
int i = 0;
int j = input.length()-1;
while(i <= input.length()/2 && j>=input.length()/2+1) {
if(input.charAt(i++) == input.charAt(j--)) {
isPalindrome = true;
} else {
isPalindrome = false;
}

}
if(isPalindrome) {
System.out.println("Palindrome");
} else {
isPalindrome = false;
System.out.println("Not a Palindrome");
}
}
}

class palindrome
{
public static void main(String args[])
{
String data="abcdcba";
int length=data.length()-1;
int i=0;
while(data.charAt(i)==data.charAt(length)&&i++<length--);
if(i>=length)
System.out.println("True");
else
System.out.println("false");
}
}

bool isPalindrome(string str)
{
for (int i = 0; i < str.size()/2; i++) {
if (str[i] != str[str.size()-i-1]) {
return false;
}
}
return true;
}

public boolean isPalindrome(String str){
	return str.equals(new StringBuilder(str).reverse().toString()); 		
}

bool Ispalindrome(string s)
{
	int i = s.size()/2 + 1;
	int j = s.size()/2 + 1;
	if (s.size()%2 == 0)
	{
		i--;
	}
	while ( i >= 0)
	{
		if (s[i]!=s[j])
			return false;
		i--;
		j++;
	}
	return true;
}

boolean isPalindrome(String str){
        if(str==null) return false;
        int length=str.length();
        int start=0;
        int end=length-1;
        str=str.toLowerCase();
        
        while(start<end){
            if(str.charAt(start)==' ') start++;
            if(str.charAt(end)==' ') end--;
            if(str.charAt(start)!=str.charAt(end)){
               return false;
            }
            start++;
            end--;
        }
        return true;

}

Can we XOR the ascii values of the entire string ?

{
{
{

for(i = 0 , j = strlen(str-1) ; i < j && str[i] == str[j] ; ++i , --j);
if(i >=j )
puts("palndrome hai yaar");

}
}
}