تحليل النصوص باستخدام الستاك: تحقق من تطابق نمط اللغة

It looks like there are a few issues in your code that might be causing it to return incorrect results for some cases. Let’s break down some of the key areas where the logic might be incorrect:

Character Comparison: The way you’re comparing characters in your stacks might not be correct. When comparing the two stacks, you should iterate through them until both stacks are empty. However, your current approach returns immediately after comparing the first pair of characters, which might not be correct for all cases.
Stack Handling: Your code uses three stacks (stack1, stack2, and stack3), but it’s not entirely clear why you need three stacks for this specific language pattern. It seems like you could achieve the same functionality with just two stacks.
Handling Non-A and Non-B Characters: Your code checks if the character is neither ‘A’ nor ‘B’ and returns false. However, it seems like you might want to ignore non-‘A’ and non-‘B’ characters and continue processing the input.
Palindromes: For the case of palindromes (e.g., ‘ABBA’), your current approach might not work correctly because you’re comparing characters from two separate stacks, which may not align correctly for palindromes.

Here’s a revised version of your code with some improvements:

java
public static boolean checkL4(File file) throws IOException {
    Stack stack1 = new Stack<>();
    Stack stack2 = new Stack<>();
    boolean bStart = false;

    try (Scanner sc = new Scanner(file).useDelimiter("\\s*")) {
        while (sc.hasNext()) {
            char w = sc.next().charAt(0);
            if (w != 'A' && w != 'B') {
                continue; // Ignore non-'A' and non-'B' characters
            }
            if (w == 'A') {
                if (!bStart) {
                    stack1.push(w);
                } else {
                    stack2.push(w);
                }
            } else if (w == 'B') {
                bStart = true;
            }
        }
    }

    while (!stack1.isEmpty() && !stack2.isEmpty()) {
        if (stack1.pop() != stack2.pop()) {
            return false; // Mismatched characters
        }
    }

    return stack1.isEmpty() && stack2.isEmpty();
}

In this revised version, I’ve simplified the stack handling to use just two stacks (stack1 and stack2). I’ve also fixed the character comparison logic to compare characters from both stacks until both stacks are empty. Additionally, I’ve added a check to ignore non-‘A’ and non-‘B’ characters in the input.

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