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Answer Posted / jaroosh
The above method is working of course, but is not the most
efficient. Other methods are however quite complex and not
so easy to explain.
Anyway, to exemplify this aforementioned method, maybe not
the most efficient code, but off the top of my head, hope
there are no misspellings.
bool isCyclic(LinkedNode *list)
{
if(list == NULL || list->next == NULL) return false;
LinkedNode *node1 = list, *node2 = node1->next;
while(node1 != node2)
{
if(node1==NULL || node2==NULL || node2->next == NULL)
return false;
node1 = node1->next;
node2= node2->next->next;
}
return true;
}
NOTE: the assumption is that for noncyclic list, the last
node has next pointer set to NULL.
| Is This Answer Correct ? | 6 Yes | 7 No |
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