从c中的字符串中删除所有重复的字符 [英] removing all repeated characters from string in c

问题描述

我想从字符串中删除所有重复的字符.例如，如果我有:

I want to remove all the repeated characters from a string. For example, if I have:

"abcdabef"

我希望结果是

"cdef"

我已经尝试过使用循环，但是这让我感到困惑.谁能告诉我该怎么做?

I have tried with loops, but it's getting me confusing. Can anyone just tell me how to do this?

这是我到目前为止尝试过的:

Here's what I've tried so far:

#include<stdio.h>
#include<string.h>
main()
{
    char s[20],ch,*p;
    int i,j,k,cnt;
    puts("enter string:");
    gets(s);
    for(i=0;s[i];i++)
    {
    ch=s[i];
    for(cnt=0,j=0;s[j];j++)
    {
            if(ch==s[j])
            cnt++;
            if(cnt>1)
            {
            for(k=0;s[k]==ch;k++)
            {
            strcpy(s+k,s+k+1);
            if(s[k]==ch)
            {k--;}
            }
            if(s[j-1]==ch)
            j--;
            }
    }
    }
    puts(s);
}

推荐答案

如果我是你，我只计算字符串中的字符并打印出在字符串中恰好出现一次的字符.

If I were you, I would just count the characters in the string and print out those which appear exactly once in the string.

char buf[BUFSIZE]; // whatever the size is

// get user input
if (!fgets(buf, sizeof buf, stdin))
    exit(EXIT_FAILURE); // couldn't fgets()

size_t len = strlen(buf);

int counts[1 << CHAR_BIT] = { 0 };

// count each character
for (size_t i = 0; i < len; i++) {
    unsigned char ch = buf[i];
    counts[ch]++;
}

// print those which are present exactly once
for (size_t i = 0; i < 1 << CHAR_BIT; i++) {
    if (counts[i] == 1) {
        printf("%c", (unsigned char)(i));
    }
}

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