如何查找数字是否是回文？

Question

Here is my code:

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

int main()
{
    unsigned int n;
    unsigned long series[100], beck[100];
    int j=0, num=0,temp=0,notflin=0;

    printf("Please enter the length_of the series :\n");

    printf("Please enter the series :\n");
    scanf("%u",&n);

    for(unsigned int f=0; f<n; f++)
    {   series[f]=0;
        scanf("%lu",&series[f]);
    }
        printf("check");
        printf("%u",n);

    for(unsigned int i=(n-1); i>=0 ; i--)
    {
        num=0;
        num=series[i];
        temp=0;
        if (num)
        {
            while((double) num/10!=0)
            {
            temp*=10;
            temp+=(num%10);
            num/=10;
            }
        }

        beck[j]=0;
        beck[j]=temp;
        j++;

    }

    unsigned int s=0,d=0;
    while(1)
    {
        unsigned long num1=0, number=0;
        number=(series[n-1-d]%10);
        num1=(beck[n-1-s]%10);
        if (number!=num1)
        {
            notflin=1;
            break;
        }
        series[n-1-d]/=10;
        beck[n-1-s]/=10;

        if(beck[n-1-s]==0)
            s++;
        if(series[n-1-d]==0)
            d++;
        if (d==(n-1)||s==(n-1))
            break;
    }

    printf(notflin==0? "Yes\n":"No\n");

    return 0;
}

I tried to enter the input to unsigned long beck[] so that the number in the last index will become the first in unsigned long series[] and then compare the numbers in the indexes.

when I try to run the code it gets stuck. What's the problem?

Answer 1

You may use the following code to find if a number is palindrome. The method returns 0 if the number is palindrome and returns -1 otherwise -

public int isPalindrome(int number) {

        int temp = number;
        int reverseNumber = 0;
        int rem = 0;

        while (temp != 0) {
            int rem = temp % 10;
            reverseNumber = reverseNumber * 10 + rem;
            temp = temp / 10;
        }

        // if 'number' and reverse of number is equal means
        if (number == reverse) {
            return 0;
        }
        return -1;
    }

Answer 2

Let's consider a case:
(input value)

As we all know the number is palindrome. Now, observe one thing, first character (from left side) is equal to first character (from right side). Similarly, second character (from left side) is equal to second character (from right side). And so on.... If this property holds for each index in the character string then the string is said to be a palindrome.

If we put a mirror in the middle of the string, the left side of the mirror will become the mirror-image of the right side, which gives the idea of optimization.

Now take two indexes : and , where points to the left-most character of the string and points to the right-most character of the string. And do as follows:

while(low <= high) {
    if(value at low index == value at high index) {
        low++;
        high--;
    } else {
        break;
    }
}

Now, if the value of is greater then the value of , means string is palindrome(HOW???). Think over it, and let me know in the comment section.

You may find this Code Snippet useful :

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

char str[100005]={0};

int main() {
    printf("Enter a number or string to be verified  :    ");
    scanf("%s",str);

    int len = strlen(str);

    int low = 0;
    int high = len -1;

    while(low <= high) {
        if(str[low] == str[high]) {
            low++;
            high--;
        } else {
            break;
        }
    }

    if(low < high) {
        printf("No !!! The input string is not palindorme ");
    } else {
        printf("Indeed !!! String is palindorme ");
    }

    return 0;
}