Program to Check whether the given string is palindrome or not .

There are two ways . One method is by using stack and second method is by comparing characters 1,n(length of the string) and 2,n-1 and so. For first method space complexity is O(n) . But for the second method  space complexity is O(1) .So it is better to use the second method. And the time complexity for both method is O(n)

Code: (in C)
int is_palindrome(char *str)
{
          if(str==NULL)
                  return 0;
          int len = strlen(str),i=0,j=len-1;
          while (i  < j)
          {
                 if(str[i]!=str[j])
                       return 0;
                 i++;
                 j--;
            }
           return 1;
}

Input   :  ababa
Output : 1

Please let me know if there is any bug in the above code and your suggestions are always welcome .

Program to Segregate odd and even nodes in linked list .

Here odd element doesn't mean that the element in the odd position . If the data is odd , it is odd element . To segregate odd and even nodes the algorithm is given below . The simplest and easiest approach is to create two linked list and when you traverse odd element node add it to odd list and add even element nodes to Even list . Otherwise you need to change links and push even nodes to the end . However that's not a good idea .

Code: (in C)

struct node* segregate_odd_and_even_nodes(struct node* head)
{
  if(head==null)
return null;
struct node* head1=null,tail1=null,head2=null,tail2=null,node=head;
while(node!=null)
{
if(node->data%2!=0)
{
if(head1==null)
{
head1=node;
tail1=node;
}
else
{
tail1->link=node;
tail1=node;
}
}
else
{
if(head2==null)
{
head2=node;
tail2=node;
}
else
{
tail2->link=node;
tail2=node;
}
}
node=node->link;
}
if(tail1!=null)
tail1->link=head2;
if(tail2!=null)
tail2->link=null;
    if(head1!=null)
            return head1;
    else
            return head2;
}

Input:      2->3->1->4->NULL
Output:    3->1->2->4->NULL
So the odd nodes will always come first in the output .

Please let me know if there is any bug in the above code and your suggestions are always welcome .

Simple C Program to Find Loop in the Linked List

This is a famous question asked in many interviews . And you probably know the answer . The Idea to solve this problem is to use two pointers, where one pointers moves by one pointer and another pointer moves by two pointers . The general idea is if two people are running in  a circle if one person is running at 1x speed and another one is running at 2x speed they will meet after 2 rounds(person who is running at 1x speed will complete one round where other one completes 2 rounds ).

Code : (in C)
int detect_loop(struct node * head)
{
if(head==null && head->link==null)
return 0;
struct node* first,second;
first=head;
second=head->link;
while(first!=second)
{
first=first->link;
if(second->link==null || second->link->link==null)
return 0;
second=second->link->link;
}
if(first==second) // found the loop
return 1;
return 0;
}

if the function returns 0 that means there is no loop , otherwise there is a loop in the linked list .

please let me know if there is any bug in the above code.