如何在C语言中仅使用数组、循环与条件语句移除字符数组中的所有重复字符

Let's break down what's wrong with your current code first, then walk through a correct implementation that meets your requirements (only arrays, loops, if/else, no pointers).

Issues with Your Current Code #

• Input Handling: Using scanf("%c", &c[i]) for 10 iterations will read whitespace (like newlines) if your input is shorter than 10 characters, which messes up your data. You also don't track the actual length of the input string.
• Duplicate Tracking Logic: Your nested loop only marks com[i] when c[i] == c[j], but this misses cases where the same character repeats in later positions (e.g., the second 'a' in "aabccdee" won't be marked). The cnt variable also counts duplicate pairs instead of unique duplicate characters, leading to incorrect results.
• Result Array Construction: The final loop that builds result has backwards logic—you're trying to exclude duplicate characters but the nested loops don't correctly filter them, leading to garbage or incorrect output.

Correct Implementation #

The core idea is to first count how many times each character appears in the input, then collect only the characters that appear exactly once. Here's how to do it:

#include <stdio.h>

int main() {
    // Define arrays with enough space (adjust if needed; 100 is safe for most cases)
    char input[100];
    int char_count[256] = {0}; // Tracks occurrence count of each ASCII character
    char result[100] = {0};
    int input_len = 0;
    int result_idx = 0;

    // Step 1: Read input string and get its length
    printf("Enter the character string: ");
    // Read until newline (avoids reading extra whitespace)
    while ((input[input_len] = getchar()) != '\n') {
        input_len++;
    }

    // Step 2: Count occurrences of each character
    for (int i = 0; i < input_len; i++) {
        // Convert char to its ASCII value to index into char_count
        char_count[(unsigned char)input[i]]++;
    }

    // Step 3: Build result array with characters that appear exactly once
    for (int i = 0; i < input_len; i++) {
        if (char_count[(unsigned char)input[i]] == 1) {
            result[result_idx] = input[i];
            result_idx++;
        }
    }

    // Step 4: Output the result
    printf("Result: %s\n", result);

    return 0;
}

How This Works #

1. Input Handling: We use getchar() to read each character until a newline, which lets us track the actual length of the input without extra whitespace.
2. Counting Characters: The char_count array (size 256 to cover all ASCII characters) keeps track of how many times each character appears. For every character in the input, we increment its corresponding index in char_count.
3. Building the Result: We loop through the input again, and only add characters to the result array if their count in char_count is exactly 1 (meaning they don't have duplicates).
4. Output: The result array is printed as a string (since we initialized it to 0, it's null-terminated correctly).

Testing this with your example input aabccdee:

• The count for 'a' is 2, 'b' is 1, 'c' is 2, 'd' is 1, 'e' is 2.
• The result array will collect 'b' and 'd', so the output is bd—exactly what you need.

If you specifically need to use fixed-size arrays (like your original 10-length arrays), you can modify the code to cap the input length at 9 (leaving space for the null terminator):

#include <stdio.h>

int main() {
    char input[10] = {0};
    int char_count[256] = {0};
    char result[10] = {0};
    int input_len = 0;
    int result_idx = 0;

    printf("Enter up to 9 characters: ");
    // Read up to 9 characters or until newline
    while (input_len < 9 && (input[input_len] = getchar()) != '\n') {
        input_len++;
    }

    // Count occurrences
    for (int i = 0; i < input_len; i++) {
        char_count[(unsigned char)input[i]]++;
    }

    // Build result
    for (int i = 0; i < input_len; i++) {
        if (char_count[(unsigned char)input[i]] == 1) {
            result[result_idx] = input[i];
            result_idx++;
        }
    }

    printf("Result: %s\n", result);

    return 0;
}

内容的提问来源于stack exchange，提问作者Hongwon Lim