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How to Enhance Palindrome Check String Algorithm
The provided code compares each character of a word to its corresponding character from the end, effectively checking for palindromes. While this approach is functional, there are optimizations that can improve its efficiency.
A better solution involves using two pointers that move towards each other from the beginning and end of the word. The following modified code addresses this:
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:
Consider the word "andna."
This modification enhances code efficiency by eliminating the loop condition that checks for even or odd word length, making it more concise and performant.
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