交换字符串中两个最频繁出现的字母

眼前的问题

strcspn（）函数将两个字符串作为参数，但是您要传递一个字符串和一个字符。您需要以某种方式将字符转换为字符串。一种方法是：

The immediate problem

The strcspn() function takes two strings as arguments, but you're passing a string and a character. You need to convert the character into a string somehow. One way to do that would be:

int sep[2] = "";
sep[0] = letter2;
string[strcspn(string, sep)] = letter3;
sep[0] = letter3;
string[strcspn(string, sep)] = letter2;

但是，第一次通话会更改 letter2 $ c的首次出现$ c>到 letter3 ；第二个调用将 letter3 （可能是上一个调用中刚替换的第一次出现）更改为 letter2 。转换字符串并不是一项完整的工作，您需要扫描整个字符串以进行更改。

However, the first call changes the first occurrence of letter2 to letter3; the second call changes the first occurrence of letter3 (which might be the one just replaced in the previous call) with letter2. This is not a complete job of transforming the strings — you need to scan the whole string making the changes.

一种可能是这样：

#include <ctype.h>
#include <stdio.h>
#include <string.h>

#define NULL_VALUE '\0'

static inline void map(char *str, int len, int c_old, int c_new)
{
    for (int i = 0; i < len; i++)
    {
        if (str[i] == c_old)
            str[i] = c_new;
    }
}

int main(void)
{
    char buffer[4096];

    printf("Enter a sentence: ");
    if (fgets(buffer, sizeof(buffer), stdin) == 0)
        return 0;
    int length = strlen(buffer);
    if (length > 0)
        buffer[--length] = '\0';

    putchar('\n');
    printf("Original [%s]\n", buffer);

    int count[256] = { 0 };
    for (int i = 0; i < length; i++)
    {
        if (isalpha((unsigned char)buffer[i]))
            count[(unsigned char)buffer[i]]++;
    }

    int max1_count = 0;
    int max2_count = 0;
    char max1_value = '\0';
    char max2_value = '\0';
    for (int i = 0; i < 256; i++)
    {
        if (count[i] > max1_count)
        {
            max2_count = max1_count;
            max2_value = max1_value;
            max1_count = count[i];
            max1_value = i;
        }
        else if (count[i] > max2_count)
        {
            max2_count = count[i];
            max2_value = i;
        }
    }

    /*
    ** Since a string is a sequence of non-null character codes followed
    ** by a null byte, it is safe to use '\0' as the temporary value in
    ** the three-step swap operation
    */
    if (max2_count > 0)
    {
        map(buffer, length, max1_value, NULL_VALUE);
        map(buffer, length, max2_value, max1_value);
        map(buffer, length, NULL_VALUE, max2_value);
    }

    printf("Revised  [%s]\n", buffer);

    return 0;
}

使用宏 NULL_VALUE的唯一原因使得三行 map（）的对称性不言而喻。

The only reason for using the macro NULL_VALUE is so that the symmetry of the three map() lines is self-evident.

我调用了程序 ccswap19 ，并且我使用了Bash这里字符串来提供数据- putchar（'\n'）; 表示输出出现在与提示符不同的一行上。如果您以交互方式运行程序，则在原始打印之前会有空白行。

I called the program ccswap19, and I used Bash 'here strings' to supply the data — the putchar('\n'); means that the output appears on a separate line from the prompt. There'd be a blank line before the 'Original' printing if you ran the program interactively.

$ ccswap19 <<< "The hidden costs of the exodus are now revealed for all to see."
Enter a sentence: 
Original [The hidden costs of the exodus are now revealed for all to see.]
Revised  [Tho hiddon cests ef tho oxedus aro new rovoalod fer all te soo.]
$ ccswap19 <<< "aaaaaaaaaaaa"
Enter a sentence: 
Original [aaaaaaaaaaaa]
Revised  [aaaaaaaaaaaa]
$ ccswap19 <<< "aaaabaaaaaaa"
Enter a sentence: 
Original [aaaabaaaaaaa]
Revised  [bbbbabbbbbbb]
$