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bool isPalindrom(const char * str)
{
   int i = 0, j = strlen(str) - 1;
   while (i < j) if (str[i++] != str[j--]) return false;
   return true;
}

reverse the string, then compare with the original string.

int isPalindrome(char *str) {
   char * end = strlen(str) - 1;
   while (str < end) {
      if (*str != *end) return -1;
   }
   return 1;
}

C# Code

private bool IsPalindrome(string inputString)
{
bool isPalindrome = false;
if (new string(inputString.Reverse<char>().ToList<char>().ToArray<char>()).Equals(inputString))
{
isPalindrome = true;
}
return isPalindrome;
}

public static boolean isPalindrome(String name){
		boolean isPalindrome = true;
		int j = name.length()-1;
		for(int i = 0; i< j; i++,j--){
			if(name.charAt(i)!= name.charAt(j)){
                	isPalindrome = false;
                    break ;
                }
		}
		return isPalindrome;
	}

Why everybody is iterating through whole String. You have to traverse 1/2 string. Start from both the side.

boolean isPalindrom=true;

for(int i=0;i<str.length()/2;i++)
{
if(str.charAt(i)!=str.charAt(str.length()-1-i))
isPalindrom=false;
}

bool palindrome( string s)
{
    string::iterator p1 = s.begin();
    string::iterator p2 = s.end();
    p2--;
    while ((*p1==*p2) && (p1<p2))
    {
        cout << "testing " << *p1 << *p2  << endl;
        p1++;
        p2--;
    }
    return (p1>=p2);
}

Java CODE:


import javax.swing.JOptionPane;
public class CheckPalindrome
{
public static void main(String[] args)
{
String s=JOptionPane.showInputDialog(null,"Enter a string ","Example",JOptionPane.QUESTION_MESSAGE);
String output="";
if(isPalindrome(s))
{
output=s+" is a palindrome!";
}
else
output=s+" is not a palindrome!";
JOptionPane.showMessageDialog(null,output,"The result",JOptionPane.INFORMATION_MESSAGE);
}
public static boolean isPalindrome(String s)
{
int low=0;
int high=s.length()-1;
while(low<high)
{
if(s.charAt(low)!=s.charAt(high))
return false;
low++;
high--;
}
return true;
}

}

bool isPalindrom(const string &s) {
    for(string<char>::const_iterator iter = s.begin(); iter <= s.size()/2; ++iter)
    {
        if( *iter != *(iter+(s.size()-(iter-s.begin()))) return false;
    }
    return true;
}
Complexity O(N/2)