@snehal ..although ur approach ir good but u make d problem little
complex,also missed out little checking, it can be done by using 2
pointers & single while loop--Now Instruction(CPU) Matters.
Rather then presenting different-2 algorithm i will present very
famous algorithm Floyd’s Cycle-Finding Algorithm:
it is the fastest method. Traverse linked list using two pointers.
Move one pointer by one and other pointer by two. If these pointers
meet at some node then there is a loop. If pointers do not meet then
linked list doesn’t have loop.
Time complexity O(n)
Space Complexity O(1)
Code-Snippet Below
int is_loop(struct node *list)
{
struct node *slow_p = list, *fast_p = list;
while(slow_p && fast_p &&
slow_p->next &&
fast_p->next &&
fast_p->next->next )
{
slow_p = slow_p->next;
fast_p = fast_p->next->next;
if (slow_p == fast_p)
{
printf("Found Loop");
return 1;
}
}
return 0;
}
It Will Surely Works.
Regards
Shashank Mani "Don't b Evil U Can Learn whilw U Earn"
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