字符串中的字符计数-C程序

Question

I have written a C program. It's a character counting program. I will give input as below

Input: ABCAPPPRC
And need as output: A2B1C2P3R1 .

But it gives output as A2B1C2A1P3P2P1R1C1 . It basically doing as per the logic I have written in program. But I don't want to count the characters of string which have already been counted. Can you suggest what logic I should implement for this?

#include <stdio.h>

int main()
{
        char str[30]= "ABCAPPPRC";
        char strOutPut[60]="";
        char *ptr= &str, *ptr2=&str;
        char ch='A';
        int count=0;
        puts(str);

        while (*ptr !=NULL)
        {
                count =0;
                ch = *ptr;
                while (*ptr2!= NULL)
                {
                        if (*ptr2 == ch) count++;
                        ptr2++;
                }
                printf("%c%d",*ptr, count);
                ptr++;
                ptr2 = ptr;
        }
}

Answer 1

You need to separate the counting from the printing.

1. The first loop goes through the input and counts the number of occurrences of each character, storing the counts in an array indexed by the character code.
2. The second loop goes through the array of counts and prints the character corresponding to a non-zero count followed by that count.

For example:

#include <stdio.h>

int main(void)
{
    char str[] = "ABCAPPPRC";
    int counts[256] = { 0 };

    puts(str);

    for (char *ptr = str; *ptr != '\0'; ptr++)
        counts[(unsigned char)*ptr]++;
    for (int i = 0; i < 256; i++)
    {
        if (counts[i] != 0)
            printf("%c%d", i, counts[i]);
    }
    putchar('\n');
    return(0);
}

Sample output:

ABCAPPPRC
A2B1C2P3R1

---

I could not understand the first for loop. Could you please explain it?

The for control line steps through the string str one character at a time. It is the for loop equivalent of the outer while loop in your original code.

char *ptr = str;
...
while (*ptr != '\0')
{
    ...
    ptr++;
}

The body of the loop converts *ptr (a plain char ) into an unsigned char (so that it is guaranteed to be positive), and then uses that value as an index into the array counts . Thus, for example, on the first iteration, A is mapped to 65, and counts[65] is incremented. Thus, for each character code, the loop increments the count corresponding to that character code each time the character is encountered in the string.

The second loop then picks out the non-zero counts, printing the character code as a character followed by its count.

(Incidentally, you should have been getting a compilation warning from the original char *ptr = &str about a type mismatch between char * and char (*)[30] . Learn when to put ampersands in front of array names — you seldom do it unless there is also a subscript after the array name. Thus, &array is usually — but not always — wrong; by contrast, &array[0] is very often valid. Also note that on some machines, NULL is defined as ((void *)0) and this elicits a warning when you compare it to a plain char , as you did with while (*ptr != NULL) . You should compare characters to '\\0' as in my rewrite; you should reserve NULL for use with pointers.)

Answer 2

str is alerady a character pointer, so when you do this: char *ptr= &str you convert a pointer to pointer to character to a char* . Loose the ampersand( & ).

Also in the inner cycle you should check if the given value of ch has already been processed. In the case you use when ptr is pointing to the second A you should just continue, because you have already added the number of A -s in the answer.

Your solution is far from optimal. I strongly suggest you lookup counting sort . It will make your solution faster but also will make it simpler.

Answer 3

@ Jonathan your solution is correct only when string characters are given in ascending order like ABCDEF, but it gives problem when character order is changed. Input string is "ABAPPPRCC" and required output is A2B1P3R1C2. Here in this case your solution will change out put to A2B1C2P3R1.

Below program gives character count without changing string formation.

char *str= "ABAPPPRCC";
    char strOutPut[30]="";
    char *ptr = str, *ptr2 = str;  
    char ch='A';
    int count=0, i = 0 , total_print = 0;
    puts(str);

    while (*ptr != '\0')
    {
            count =0;
            ch = *ptr;
            while (*ptr2!= '\0')
            {
                    if (*ptr2 == ch) count++;
                    ptr2++;
            }
            for( i = 0; i < total_print ; i++ )
            {
                if ( ch == strOutPut[i] )
                {
                     i =  total_print + 1;
                       break;
                }
            }  

            if( i <= total_print )
            {
                    printf("%c%d",*ptr, count);
                    strOutPut[total_print++] = ch;
            }
            ptr++;
            ptr2 = ptr;
    }

Answer 4

#include <stdio.h>

int main(void){
    const char noncountchar = '\x11';
    char str[30]= "ABCAPPPRC";
    char strOutPut[60]="";
    char *ptr, *ptr2;
    char ch;
    int count=0, len=0;
    puts(str);

    for(ptr=str;ch=*ptr;++ptr){
        if(ch == noncountchar) continue;
        count =1;
        for(ptr2=ptr+1;*ptr2;++ptr2){
            if (*ptr2 == ch){
                *ptr2 = noncountchar;
                ++count;
            }
        }
        len += sprintf(strOutPut+len, "%c%d", *ptr, count);
    }
    printf("%s", strOutPut);
    return 0;
}