基于C++的游程编码解压实现求助

Hey there! Let's get that decompression function working properly for your run-length encoding (RLE) setup. First, let's recap how your encoder works: it replaces consecutive repeated characters with a count followed by the character, and leaves single characters (including spaces!) as-is. Your decompressor needs to reverse this logic, and we'll fix your messy code step by step.

First, Let's Diagnose the Issues in Your Current Code #

Your existing decompress function has a few key problems:

• You're comparing characters directly to integers (like compressed[i] <= 1)—this checks the ASCII value of the character, not the numeric value, which is totally wrong.
• You're jumping ahead with i++ too early, which leads to skipping characters or incorrect indexing.
• You aren't handling multi-digit counts (though your example uses single digits, your encoder would produce things like "10a" for 10 repeated 'a's, so we need to account for that).
• The logic flow is reversed: you should first check if a character is a digit (part of a count) before treating it as a single character.

The Corrected Decompression Code #

Here's a working implementation that handles all cases, including multi-digit counts and spaces:

#include <string>
#include <cctype> // Required for isdigit()

void Decompress(std::string &compressed, std::string &original) {
    original.clear(); // Ensure we start with an empty result string
    int n = compressed.length();
    int i = 0;

    while (i < n) {
        // Case 1: We've hit a digit (start of a count)
        if (isdigit(compressed[i])) {
            std::string countStr;
            // Collect all consecutive digits to form the full count (handles 10+, 100+, etc.)
            while (i < n && isdigit(compressed[i])) {
                countStr += compressed[i];
                i++;
            }
            // Convert the digit string to an integer
            int count = std::stoi(countStr);
            // The next character is the one we need to repeat
            if (i < n) {
                char targetChar = compressed[i];
                // Append the character 'count' times efficiently
                original.append(count, targetChar);
                i++;
            }
        } 
        // Case 2: Non-digit character (single occurrence, like spaces or single letters)
        else {
            original += compressed[i];
            i++;
        }
    }
}

How This Works #

Let's break down the key parts:

1. Clear the result string: We start by clearing original to avoid appending to any existing content.
2. Loop through the compressed string: We use a while loop instead of a for loop to give us more control over incrementing i (since we might jump multiple positions when collecting digits).
3. Handle digits: When we hit a digit, we collect all consecutive digits to form the full count (so "123a" becomes a count of 123). We convert this string to an integer with std::stoi.
4. Repeat the character: Using string.append(count, targetChar) is a clean, efficient way to add the repeated character to the result.
5. Handle single characters: Any non-digit (like spaces, 'i', 'k', etc.) is added directly to the result, since they represent single occurrences from the original string.

Testing with Your Example #

If you pass your compressed string "3a4h3i 3k2j2h ikl 6w4e2t" into this function, it will correctly output the original string: "aaahhhhiii kkkjjhh ikl wwwwwweeeett".

内容的提问来源于stack exchange，提问作者liasis