Leetcode. Linked list cycle && Linked list Cycle II

Given A linked list, determine if it has a cycle in it.

Follow up:
Can you solve it without using extra space?

With two pointers fast and slow,fast each step, slow each step, if fast catch up with slow there is a ring, otherwise does not exist.

1 BOOLHascycle (ListNode *head)2     {3ListNode *slow = head, *fast =head;4          while(Fast! =NULL)5         {6slow = slow->Next;7Fast = Fast->Next;8             if(Fast! =NULL)9Fast = Fast->Next;Ten             if(Fast! = NULL && Fast = =slow) One                 return true; A         } -          -         return false; the}

The Linked List Cycle II is the node that asks for the ring to appear.

When fast catches up with slow, fast is more than slow go nc,n>=1, C is the circumference of the circle. Such as:

Slow walking distance: LenA + x,fast walk: LenA + NC + x

Because Fast walk is twice times the distance of Slow: 2 (LenA + x) = LenA + NC + x

Thus, NC = LenA + x, i.e. LenA = Nc-x.

Then meet, let slow point to head,fast and slow walk the same step, and finally meet in the join place.

1ListNode *detectcycle (ListNode *head)2     {3ListNode *fast = head, *slow =head;4         5          while(Fast! =NULL)6         {7slow = slow->Next;8Fast = Fast->Next;9             if(Fast! =NULL)TenFast = Fast->Next; One             if(Fast! = NULL && Fast = =slow) A                  Break; -         } -          the         if(Fast = =NULL) -             returnNULL; -              -slow =head; +          while(Slow! =fast) -         { +slow = slow->Next; AFast = Fast->Next; at         } -          -         returnslow; -}

Leetcode. Linked list cycle && Linked list Cycle II