我正在尝试比较字符串文字，我想删除重复的文字，我想在不使用 POINTERS 的情况下进行。，

Question

I am trying to compare string literals and I want to remove repeated literals I want to do it without using POINTERS.

This is my code:

char str[30];

printf("Enter strings : ");     

fgets(str,29,stdin);

char tem[30];

int count , county;

for(count = 0 ; count < strlen(str)-1 ; count++) {      
    for(county = 1 ; county < strlen(str) ; county++) {
        if(str[count] != str[county]) {
               tem[count] = str[count];     
        }
    } 
}

//PRINT

for(count = 0 ;count < strlen(str) -1 ; count++) {

    printf("%c",tem[count]);

}

Input: happen
Expected output: hapen

Answer 1

Correcting your code:

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

int main()
{
    char str[30];

    printf("Enter strings : ");

    fgets(str,30,stdin);

    char tem[30];

    size_t count;
    size_t county=0;;

    for(count = 0 ; count < strlen(str)-1 ; count++) {
            if(str[count] != str[count+1]) {
                   tem[county++] = str[count];
            }
    }

    tem[county] = '\0';

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

    return 0;
}

Take note that this code remove double chars if this char have +1 displacement in the sting.

EDIT

To have mspi as output of entered string mississippi

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

int main()
{
    char str[30];

    printf("Enter strings : ");

    fgets(str,30,stdin);

    char tem[30];

    size_t count;
    size_t county;
    size_t tem_index = 0;
    size_t size_of_string = strlen(str);
    bool found;

    for(count = 0 ; count < size_of_string-1; count++)
    {
        found = false;
        county = count+1;

        while ((found == false) && (county<size_of_string))
        {
            if(str[count] == str[county])
            {
                found = true;
            }
            county++;
        }

        if (found == false)
        {
            tem[tem_index++] = str[count];
        }
    }

    tem[tem_index] = '\0';

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

    return 0;
}

EDIT 2

To have misp as output of entered string mississippi

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

int main()
{
    char str[30];

    printf("Enter strings : ");

    fgets(str,30,stdin);

    char tem[30];

    size_t count;
    size_t county;
    size_t tem_index = 0;
    size_t size_of_string = strlen(str);
    bool found;

    for(count = 0 ; count < size_of_string-1; count++)
    {
        found = false;
        county = 0;

        while ((found == false) && (county<tem_index))
        {
            if(str[count] == tem[county])
            {
                found = true;
            }
            county++;
        }

        if (found == false)
        {
            tem[tem_index++] = str[count];
        }
    }

    tem[tem_index] = '\0';

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

    return 0;
}

Take note that all these solutions are case sensitive.

Answer 2

Your code to remove allduplicates isn't quite there yet:

• You use the same indices for the new and the old string, but you need two different indices, because the new string is as long or shorter than the old string. If the old string is "aaab", your index for the old string is 3 when you see the "b", but the index for the new string is only 1. (By skipping indices you leave uninitialised gaps in your string.)

• You look forward to find other occurrences of the same letter, but you append to the new string for every letter that doesn't atch. You must look at all folllowing letters, but you must append to the new string only once. That is, you must make your decision whether the letter is duplicate or not after the loop, based on the information that you've found in the loop.

• When you look forward, you shouldn't start a 1, but at the letter after he current letter. If you start at one, you will find duplicates for every letter after the first, because you check each letter with itself.

• This is not an arror, but it's not a good idea to call strlen repeatedly in a loop. The length of the input string doesn't change, so you can determine the string length beforehand. If you just want to use it as your termination condition, you can test whethet the current letter is the null terminator.

Below is a solution that uses your logic, albeit by looking backwards, not forward. (If you look forward, you will copy the last occurence of a letter, if you look back, you'll copy the first occurrence. It may make a difference in the order of the letters. For Mississippi, you'll get "Mspi" or "Misp" depending on which strategy you use.)

The program overwrites the same string. This is possible, because you are filtering out letters and the new index is equal to the old index or smaller:

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

void remdup(char *str)
{
    int i = 0;      // index into old string
    int j = 0;      // index into new string

    for (i = 0; str[i]; i++) {
        int k = 0;
        int dup = 0;

        for (k = 0; k < i; k++) {
            if (str[i] == str[k]) {
                dup = 1;
                break;
            }
        }

        if (dup == 0) str[j++] = str[i];
    }

    str[j] = '\0';
}

int main()
{
    char str[] = "Mississippi";

    puts(str);
    remdup(str);
    puts(str);

    return 0;
}

This solution doesn't scale for large strings. A more effective method would be to keep a table of which of the 256 possible characters have already been used.