交换字符串中出现频率最高的两个字母

Question

我不知道我的代码有什么问题，但是当我编译时我得到：

warning: passing arg 2 of `strcspn' makes pointer from integer without a cast

代码如下：

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

#define STR_LEN 50

int main(void) {
    int i = 0, j = 0, length = 0, count1 = 0, count2 = 0, count3 = 0;
    char letter3 = 'a', letter2 = 'a', string[STR_LEN] = { 0 };

    length = strlen(string);
    printf("Enter a sentence: ");
    fgets(string, STR_LEN, stdin);

    for (i = 0; i < length; i++) {
        for (j = 0; j < length; j++) {
            if (string[i] == string[j]) {
                count1++;
            } else {
                count1 = 0;
            }
        }
        if (count1 > count3) {
            count2 = count3;
            count3 = count1;
            letter2 = letter3;
            letter3 = string[i];
        } else
        if (count1 > count2) {
            count2 = count1;
            letter2 = string[i];
        }
    }

    string[strcspn(string, letter2)] = letter3;
    string[strcspn(string, letter3)] = letter2;

    printf("\n %s", string);

    system("pause");

    return 0;
}

代码应该从用户那里得到一个句子，并将句子中最常见的字母与第二个常见的字母切换。

Answer 1

strcspn(const char *str1, const char *str2) 在第一个字符串中查找第二个字符串中任何字符的第一个实例。您正在传递一个字符而不是字符串作为第二个参数。您需要查找单个字符的函数 strchr(const char *string, int character)。

Answer 2

眼前的问题

strcspn() 函数接受两个字符串作为参数，但您传递的是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：

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

但是，第一次调用将 letter2 的第一次出现更改为 letter3；第二次调用将第一次出现的 letter3（可能是上次调用中刚刚替换的那个）更改为 letter2。这不是 t运行形成字符串的完整工作 — 您需要扫描整个字符串进行更改。

实施解决方案

一种可能是这样的：

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

#define NULL_VALUE '[=11=]'

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] = '[=11=]';

    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 = '[=11=]';
    char max2_value = '[=11=]';
    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 '[=11=]' 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 的唯一原因是三个 map() 线的对称性是 self-evident.

示例运行

我调用了程序 ccswap19，我使用 Bash 'here strings' 来提供数据 — putchar('\n'); 意味着输出与迅速的。如果您运行程序是交互式的，那么在 'Original' 打印之前会有一个空行。

$ 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]
$

交换字符串中出现频率最高的两个字母

Swap occurrences of two most frequent letters in a string

c

string

warnings

compiler-warnings

string.h

眼前的问题

实施解决方案

示例运行