Python3中如何螺旋遍历单词字符并返回布尔值？

Hey there! I get that trying to handle spiral traversal (grabbing characters from outside in, alternating first and last) with fixed index tricks like [0] or [:-1] can get messy when you need to support words of any length. Split isn't really the right tool here—instead, let's use a two-pointer technique that dynamically tracks the start and end of our remaining character range. It's clean, adaptable, and easy to understand.

Step 1: Implement Spiral Traversal #

The core idea is to use two pointers: one starting at the very beginning (left) of the word, and one at the end (right). We'll alternate adding characters from these pointers to our result list, moving the pointers inward each time until they cross each other. This works seamlessly for both even and odd-length words.

def get_spiral_sequence(word):
    char_list = list(word)
    left = 0
    right = len(char_list) - 1
    spiral = []
    
    while left <= right:
        # Add character from the left pointer
        spiral.append(char_list[left])
        left += 1
        
        # Avoid duplicate middle character if word length is odd
        if left > right:
            break
        
        # Add character from the right pointer
        spiral.append(char_list[right])
        right -= 1
    
    return spiral

Let's test this with your example word 'urgent':

print(get_spiral_sequence('urgent'))  # Output: ['u', 't', 'r', 'n', 'g', 'e']

Perfect—this matches the traversal order you described.

Step 2: Check if the Spiral Sequence is Alphabetical #

From your example, we need to return False when the spiral sequence is in reverse alphabetical order. Let's write a helper function to verify if the sequence follows ascending alphabetical order (returning True if it does, False otherwise).

def is_alphabetical(sequence):
    # Iterate through consecutive character pairs
    for i in range(len(sequence) - 1):
        # If current character is >= next one, it's not in order
        if sequence[i] >= sequence[i + 1]:
            return False
    return True

Step 3: Combine into One Function #

Putting it all together, here's a single function that handles both the spiral traversal and the order check:

def check_spiral_alphabetical(word):
    # Generate the spiral sequence
    char_list = list(word)
    left, right = 0, len(char_list) - 1
    spiral = []
    
    while left <= right:
        spiral.append(char_list[left])
        left += 1
        if left > right:
            break
        spiral.append(char_list[right])
        right -= 1
    
    # Verify alphabetical order
    for i in range(len(spiral) - 1):
        if spiral[i] >= spiral[i + 1]:
            return False
    return True

Testing with your example:

print(check_spiral_alphabetical('urgent'))  # Output: False

This aligns with your expected result, since the spiral sequence is reverse alphabetical.

Why This Works Better Than Fixed Indexes #

• Adapts to any length: Whether your word has 3 characters or 30, the two pointers adjust dynamically—no hardcoded indexes to break things.
• Avoids duplicates: The if left > right: break check ensures we don't add the middle character twice for odd-length words.
• Easy to modify: If you ever need to check for reverse alphabetical order instead, just flip the condition to if sequence[i] <= sequence[i + 1]: return False.

内容的提问来源于stack exchange，提问作者jess