找出每个字符在字符串链表中出现的次数[重复]

Question

我必须找出每个字符在字符串链表中重复的次数。字符串被存储并从文件中读取。我必须以两种方式打印结果：字母顺序和增长顺序。

我试图编写一个函数来计算给定字符重复的次数，但它崩溃了。

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


struct list {
    char *string;
    struct list *next;
};

typedef struct list LIST;

int count(struct list* head, char search) // funct to calculate how many 
                                          //times 1 string appears
{
    struct list* current = head;
    int count=0;
    while (current!=NULL)
    {
        if(current->string == search)
            count++;
    }
    return count;
}

int main(void) {
    FILE *fp;
    char line[128];
    LIST *current, *head;

    head = current = NULL;
    fp = fopen("test.txt", "r");

    while(fgets(line, sizeof(line), fp)){
        LIST *node = malloc(sizeof(LIST));
        node->string = strdup(line);
        node->next =NULL;

        if(head == NULL){
            current = head = node;
        } else {
            current = current->next = node;
        }
    }
    fclose(fp);
    //test print
    for(current = head; current ; current=current->next){
        printf("%s", current->string);
    }

    count(head, "a");

    return 0;
}

test.txt 文件包含：

Astazi nu este maine

Answer 1

问题在于if(current->string == search) 将指针 (char *) 与 char 进行比较。如果 current->string 是单个字符，您可以使用 if(*current->string == search)。如果字符串包含多个字符，您必须告诉我字符串“aa”的count() 是什么，搜索“a”。另一个主要问题是count() 中的while 循环不会遍历链表，因此会导致无限循环。

int count(struct list *head, char search) {
    int count = 0;
    for(struct list* current = head; current; current = current->next) {
        for(int i = 0; current->string[i]; i++)
            if(current->string[i] == search) count++;
    }
    return count;
}