Check string for palindrome

A concise version, that doesn't involve (inefficiently) initializing a bunch of objects:

boolean isPalindrome(String str) {    
    int n = str.length();
    for( int i = 0; i < n/2; i++ )
        if (str.charAt(i) != str.charAt(n-i-1)) return false;
    return true;    
}

Why not just:

public static boolean istPalindrom(char[] word){
    int i1 = 0;
    int i2 = word.length - 1;
    while (i2 > i1) {
        if (word[i1] != word[i2]) {
            return false;
        }
        ++i1;
        --i2;
    }
    return true;
}

Example:

Input is "andna".
i1 will be 0 and i2 will be 4.

First loop iteration we will compare word[0] and word[4]. They're equal, so we increment i1 (it's now 1) and decrement i2 (it's now 3).
So we then compare the n's. They're equal, so we increment i1 (it's now 2) and decrement i2 (it's 2).
Now i1 and i2 are equal (they're both 2), so the condition for the while loop is no longer true so the loop terminates and we return true.

You can check if a string is a palindrome by comparing it to the reverse of itself:

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

or for versions of Java earlier than 1.5,

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

EDIT: @FernandoPelliccioni provided a very thorough analysis of the efficiency (or lack thereof) of this solution, both in terms of time and space. If you're interested in the computational complexity of this and other possible solutions to this question, please read it!