How Can We Optimize Palindrome String Checking?

Improving Code for Palindrome Checking

To check if a string is a palindrome, you've developed a method that compares characters from both ends towards the middle. However, there's room for optimization.

Optimized Approach:

The following code offers a simpler and more efficient solution:

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;
}

Explanation:

This code utilizes two pointers, i1 and i2, initialized to the beginning and end of the word, respectively. It iterates inward until i2 crosses i1.

• If the characters at word[i1] and word[i2] don't match, the method returns false.
• If they match, both pointers move towards the center: i1 increments, and i2 decrements.

When the pointers intersect, the entire word has been successfully compared, and the method returns true.

Example:

For the word "andna":

• Initially, i1 is 0 (beginning of the word) and i2 is 4 (end of the word).

• Loop iteration:

  • Character at i1 (0) is 'a'. Character at i2 (4) is 'a'. They match, so move both pointers towards the middle.
  • Character at i1 (1) is 'n'. Character at i2 (3) is 'n'. They match.
  • i1 (2) and i2 (2) have intersected, so all characters have been compared successfully.

The method returns true, confirming that "andna" is a palindrome.

The above is the detailed content of How Can We Optimize Palindrome String Checking?. For more information, please follow other related articles on the PHP Chinese website!