Python：如何通过仅返回前10个前缀匹配元素的API获取完整列表

Great question! The key here is to exploit the API's behavior—returning up to 10 matches for a given prefix—and systematically explore all possible prefixes to ensure every element in lst is captured. Here's a step-by-step approach:

Core Idea #

The API returns all elements starting with your input prefix, capped at 10. If an API call returns exactly 10 results, it means there might be more elements with that prefix. We can use this clue to iteratively narrow down prefixes until we've collected every unique element.

Step-by-Step Implementation #

1. Initialize tracking structures:

   • A set collected to store unique elements (avoids duplicates from overlapping prefixes).
   • A queue prefix_queue to keep track of prefixes we need to check, starting with the empty string (""—which matches every element in the list).

2. Process each prefix:

   • While the queue isn't empty, pull a prefix and call the API with it.
   • Add all returned results to collected.
   • If the API returns exactly 10 results, generate new prefixes by extending the current prefix with the next character from each result (e.g., if current prefix is "b" and a result is "banana", the new prefix is "ba"). Add these unique new prefixes to the queue.

3. Finalize the full list:

   • Once the queue is empty, convert the collected set to a list—this is your complete lst.

Example Pseudocode #

collected = set()
prefix_queue = {""}  # Start with empty string to match all elements

while prefix_queue:
    current_prefix = prefix_queue.pop()
    # Call the API with the current prefix
    matches = api(current_prefix)
    # Add all matches to our collected set
    collected.update(matches)
    # If we hit the 10-match limit, there might be more elements to find
    if len(matches) == 10:
        # Generate new prefixes by taking each match's next character
        new_prefixes = {item[:len(current_prefix) + 1] for item in matches}
        # Add new prefixes to the queue (avoid reprocessing existing ones)
        prefix_queue.update(new_prefixes - prefix_queue)

# Convert the set to a list to get the full lst
full_lst = list(collected)

Why This Works #

• Every element in lst will match at least one prefix (itself, or any shorter prefix derived from it).
• When an API call returns 10 results, we know there could be more elements under that prefix—so we dig deeper with more specific prefixes to capture the rest.
• Using a set ensures we don't count duplicate elements that might be returned by multiple overlapping prefixes.

内容的提问来源于stack exchange，提问作者bb5kb