Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
1.0k views
in Technique[技术] by (71.8m points)

arrays - How to remove duplicate char in string in C

Total of Each Character

Due to the long quarantine, Jojo is getting super bored. To get rid of his boredom, Jojo proposes a game to play with Bibi. The game calculates the total count of each character in the word. Since Jojo is so eager to win, he asks you to help him guess the result of the game.

Format Input

The input starts with an integer T, the number of test cases. Every test case starts with a line containing a single string W, which indicates the word. It is guaranteed that the word will only contain alphanumeric and space.

Format Output

Every test case will start with a line of "Case #X:", where X represents the test case. Following this line is the result of each character in the word and total of the character number. The result will be sorted by alphabet first and follow with a number.

Constraints

1 ≤ |??| ≤ 10000

Sample Input

3
Bina Nusantara University
Binus 2020
Corona Virus Covid 19

Sample Output

Case #1:

[A] => 4
[B] => 1
[E] => 1
[I] => 3
[N] => 4
[R] => 2
[S] => 2
[T] => 2
[U] => 2
[V] => 1
[Y] => 1

Case #2:

[B] => 1
[I] => 1
[N] => 1
[S] => 1
[U] => 1
[0] => 2
[2] => 2

Case #3:

[A] => 1
[C] => 2
[D] => 1
[I] => 2
[N] => 1
[O] => 3
[R] => 2
[S] => 1
[U] => 1
[V] => 2
[1] => 1
[9] => 1

What I have tried:

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

int main(){
    
    int tc, count=0, len;
    char str[500][500];
    
    scanf("%d", &tc); getchar();
    for(int i = 0; i < tc; i++){
        scanf("%[^
]", str[i]); getchar();
        len = strlen(str[i]);
        printf("Case #%d:
",i+1);
        for(int j = 0; j < len; j++){
            for(int k = 0; k < len; k++){
                if(str[i][j] >= 97 && str[i][j] <= 122) str[i][j]-=32;
                if(str[i][j] == 32) continue;
                if(str[i][j] == str[i][k]) count++;
                // remove duplicate char
                if(str[i][j] == str[i][k]){
                    
                }
            }
            printf("[%c] => %d
", str[i][j], count);
            count=0;
        }
        printf("
");
    }
        
    return 0;
}
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

One of the most basic approaches when you need a count of the number of times any element within a collection occurs for a given set is to use a Frequency Array. Very simply, a Frequency Array contains the number of elements that can occur for a collection. (e.g. with what frequency does the thing mapped to each index occur?)

For example, if you want to know the number of times any one alpha-character occurs within a line of text (ignoring case), your collection has 26-elements (the letters in the alphabet). So you would simply declare an array of 26-elements (usually int or size_t), initialized all zero, and then loop over the characters in your line of text, and for each alpha-character, you increment the corresponding element of the array.

For example, with the array int chars[26] = {0}; and for the line containing "Dog", you would simply increment:

chars[toupper(line[0]) - 'A']++;   /* increments index chars[3]++  */
chars[toupper(line[1]) - 'A']++;   /* increments index chars[14]++ */
chars[toupper(line[2]) - 'A']++;   /* increments index chars[6]++  */

(see ASCII Table and Description to understand how toupper(line[0]) - 'A' maps the character to the corresponding index. Also note cast to unsigned char intentionally omitted for clarity)

After looping over all character in the line, you can output the results by looping over all elements of the frequency array, and if the value at an index is greater than zero, you output the character corresponding to that index and the value held as the count, e.g.

for (int i = 0; i < 26; i++)
    if (chars[i])
        printf ("[%c] => %d
", i + 'A', chars[i]);

To handle the digit-characters in the line, you simply use a second 10-element frequency array and do the same thing, but this time subtracting '0' to map the ASCII digits to elements 0-9 of the array.

The other aspect of the problem you have omitted is validating every input and conversion. You cannot use any input function correct unless you check the return to determine whether the operation succeeded or failed. The same applies to every conversion to a numeric type and any other expression critical to the continued defined operation of your code.

Putting the pieces together, you could use a global enum to define needed constants for your program (or to the same thing with multiple #define statements) and open your file and read and convert the first line to an integer value with:

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

enum { DIGITS=10, CHARS=26, MAXC=1024 };    /* constants for use in program */

int main (int argc, char **argv) {
    
    char line[MAXC];        /* buffer to hold each line of input */
    int ncases = 0;         /* integer for number of cases */
    /* use filename provided as 1st argument (stdin by default) */
    FILE *fp = argc > 1 ? fopen (argv[1], "r") : stdin;
    
    if (!fp) {  /* validate file open for reading */
        perror ("file open failed");
        return 1;
    }
    
    if (!fgets (line, MAXC, fp)) {              /* read/validate 1st line */
        fputs ("(user canceled input)
", stdout);
        return 0;
    }
    if (sscanf (line, "%d", &ncases) != 1) {    /* convert/validate to int */
        fputs ("error: invalid integer input for ncases.
", stderr);
        return 1;
    }
    

Now with the number of cases in ncases, you will simply loop that number of times, reading the next line from the file, looping over each character in line incrementing the corresponding elements of two frequency arrays chars and digits to count the occurrences of each character type (alpha and digits) and then output the results looping over each frequency array in sequence to output the counts from each. You can do that as follows:

    for (int i = 0; i < ncases; i++) {      /* loop ncases times */
        int chars[CHARS] = {0},             /* frequency array for characters */
            digits[DIGITS] = {0};           /* frequency array for digits */
        
        if (!fgets(line, MAXC, fp)) {       /* read/validate line */
            fputs ("error: reading line from file.
", stderr);
            return 1;
        }
        
        for (int j = 0; line[j]; j++) {     /* loop over each char in line */
            int c = toupper((unsigned char)line[j]);    /* convert to uppercase */
            if (isalpha((unsigned char)c))          /* check if A-Z */
                chars[c-'A']++;                     /* increment chars at index */
            else if (isdigit((unsigned char)c))     /* check if 0-9 */
                digits[c-'0']++;                    /* increment digits at index */
        }
        
        printf ("
Case #%d:

", i + 1);          /* output case no. */
        for (int j = 0; j < CHARS; j++)             /* loop over chars array */
            if (chars[j])                           /* if value at index non-zero */
                printf ("[%c] => %d
", j + 'A', chars[j]); /* output count */
        for (int j = 0; j < DIGITS; j++)            /* loop over digits array */
            if (digits[j])                          /* if value at index non-zero */
                printf ("[%d] => %d
", j, digits[j]);  /* output count */
    }

(note: for outputting the digits counts, there is no need to map the indexes back to their ASCII value -- you can simply output the integer representation with %d instead of mapping j + '0' to output the ASCII chars with %c -- your choice really. Also see man 3 isalpha for the requirement that each argument to any of the classification macros be of the type unsigned char -- explaining the purpose of the casts above to (unsigned char))

That is essentially the complete code for approaching your problem with frequency arrays. All you need is to tidy up and return success to the shell, e.g.

    if (fp != stdin)   /* close file if not stdin */
        fclose (fp);
    
    return 0;
}

Example Use/Output

Placing your sample input in the file dat/charcountlines.txt and then providing that as input to the program, would result in the following:

$ ./bin/charcount dat/charcountlines.txt

Case #1:

[A] => 4
[B] => 1
[E] => 1
[I] => 3
[N] => 4
[R] => 2
[S] => 2
[T] => 2
[U] => 2
[V] => 1
[Y] => 1

Case #2:

[B] => 1
[I] => 1
[N] => 1
[S] => 1
[U] => 1
[0] => 2
[2] => 2

Case #3:

[A] => 1
[C] => 2
[D] => 1
[I] => 2
[N] => 1
[O] => 3
[R] => 2
[S] => 1
[U] => 1
[V] => 2
[1] => 1
[9] => 1

The above matches the output you specify and the ordering. Look things over and let me know if you have further questions.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

1.4m articles

1.4m replys

5 comments

57.0k users

...