c - 挤压c中的单词——C语言基础

Question

我不确定我在下面的代码中做错了什么，

这就是我想要的

User Input: aaaaaadsssssc

Output: a6ds5c

我的程序一直运行良好，直到今天早上我才愚蠢到覆盖整个文件。哦，好吧，欢迎来到 Linux!由于我重写了程序，它给我段错误，我不知道我在这里做错了什么，此时我的头脑停止了工作。

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

int main(){
        int i, j, k, count;
        char str[1000];

        printf("Enter: ");
        scanf("%[^\n]s", str);
        printf("You have typed: %s\n", str);

        printf("length: %d\n", strlen(str));
        for (i = 1; i < strlen(str); i++){
                count = 2;

                if (str[i] == str[i-1]){
                        j = i;
                        while (str[i] == str[i+1]){
                                if (k != strlen(str)){str[k] = str[k+1];}
                                else{str[k] = '\0';}
                                i++; k++; count++;
                        }
                str[j] = count;
                k++;
                }
        }
        str[k] = '\0';
        printf("output: %s\n", str);
}

几天后我就要到期了，但我搞砸了。有人可以指出我犯的错误吗？非常感谢!

Answer 1

这会起作用:

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

    int main(){
            int i, count;
            char str[1000];

            printf("Enter: ");
            scanf("%s", str);
            printf("You have typed: %s\n", str);

            //used strlen() only once because strlen() takes O(n) time
            int len = strlen(str);
            printf("length: %d\n", len);

            //iterate through the string
            for (i = 0; i < len - 1; i++){

                    //every char appears at least once
                    count = 1;

                    //print the char regardless of it repeating
                    printf("%c", str[i]);
                    for (int j = i; str[j] == str[j+1]; ++j, ++i)
                            ++count;

                    //print the count if it's greater than 1
                    if (count > 1) printf("%d", count);
            }

            //print the last char if it's different from the second last
            if (str[len - 1] != str[len - 2]) printf("%c", str[len - 1]);

            //print new line
            printf("\n");
    }