Floyd's Cycle removal

#1

let’s assume a pointer p which contain the prev node of fast so our solution is as follows
if(fast==slow){
slow->next=head;
slow=slow->next;
node *p;
p->next=fast;
fast=fast->next;
if(fast==slow){
p->next=NULL;
}
is this approach correct?

#2

@Vivek-Pandey-2129725577345937

You should always mention the context (Timing in the video, or the problem statement, or some details) of the query.

The correct approach would be

node*p = head;
while(p->next != fast) p = p->next; // Getting to the previous node of fast
s = head;

while(fast!=slow){
slow=slow->next;
p = fast;
fast = fast->next
}
p->next = null;

Let me know if you till need any help.

#3

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