C递归检查回文

Question

I've got an assignment where I need to check if a given string is a palindrome or not recursively.

EG :

aD1Da // returns 1
asd1s // returns 0

The function signature needs to look like this:

int isPolindrom(char str[]);

Which means I'm getting only one parameter. I managed to build a solution using a static variable (The function will be called more than once, so we were instructed not to use static variables, though my solution works for more than 1 call).

My solution:

static flag = 0; 
int isPolindrom(char str[])
{
    static int pos = 1;
    if (flag == 1) { 
        pos = 1; // Reset
        flag = 0; // Reset
    }
    if (pos >= strlen(str)) { 
        flag = 1; 
        return 1; 
    }
    if (*str == *(str + strlen(str) - pos)) { 
        pos++; 
        return  isPolindrom(str + 1); 
    }
    else return 0; 
}

It's also possible using a secondary function, but I was wondering if and how is it possible to do this without static variables and without writing secondary functions.

In the automatic tests they are sending me the string as a constant string. For example:

isPolindrom("abcba")

Answer 1

Yes it is possible. Assuming that string is null terminated and is mutable we will do this

int isPolindrom(char s[])
{

    int l = 0;
    while(s[l]) l++;
    if(l == 2)
       return s[0]==s[1];
    if(l <= 1)
       return 1;

    char c=s[l-1];
    s[l-1]=0;

    int result = (s[0]==c) && isPolindrom(s+1);

    s[l-2]=c;
    return result ;
}

Though it doesn't change the passed argument - we are modifying it internally. That's why modifiable sting requirement was needed.

After edit:

Note my earlier answer was considering that you couldn't use any function but as OP made it clear we can easily use standard library functions like malloc etc. That's why this solution using dynamically allocated memory.

With the given constraint

int isPolindrom(const char *s)
{
    int l = 0;
    while(s[l]) l++;
    if(l == 2)
      return s[0]==s[1];
    if(l == 1)
      return 1;

    char *ss =  malloc(l-1);
    if( ss == NULL ){
       perror("Error in malloc");
       exit(EXIT_FAILURE);
    }
    for(int i=0;i<l-2;i++)
    ss[i]=s[i+1];
    ss[l-2]=0;
    int p = (s[0]==s[l-1]) && isPolindrom(ss);

    free(ss);
    return p;
}

Answer 2

Creating a new string inside palindrome() .

int palindrome(char *str)
{ 
    int len= strlen(str);
    if(len<=1)
      return true;
    char a=str[0];
    char b=str[len-1];
    char *newstr=malloc(len-2+1);
    for(int i=1;i<len-1;i++)
        newstr[i-1] = str[i];
    newstr[len-2]='\0';

    int res = (a==b && palindrome(newstr));
    free(newstr);
    return res;
}

and call it as

 palin("abcba");

Answer 3

This is my take on this:

1. check if string must be palindrome (ie less than 2 chars), if yes - return true
2. if no, check if first == last , if yes - recursive call with the inner string

---

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

bool is_palindrome(char * str){
    // end conditions
    if (!str)
        return false;
    if (strlen(str) < 2)
        return true;

    // check first vs. last
    if (str[0] == str[strlen(str) - 1]){
        // remove first & last chars (copy to array to allow this)
        char tmp_arr[strlen(str) - 1];
        strncpy(tmp_arr, str+1, strlen(str) - 2);
        tmp_arr[strlen(str) - 2] = '\0';
        return is_palindrome(tmp_arr);

        /* if string was mutable - this would be easier:

                str[strlen(str) - 1] = '\0';
                return is_palindrome(str + 1);
        */
    }

    return false;
}

int main(){
    printf(is_palindrome("abcddcba") ? "true\n" : "false\n");
    printf(is_palindrome("abcddcbaa") ? "true\n" : "false\n");
    printf(is_palindrome("abcd   dcba") ? "true\n" : "false\n");
    while(1);
    return 0;
}