O(1) solution using next pointer's loop


  • 0
    S
    public boolean hasCycle(ListNode head) {
        if(head == null || head.next == null){
            return false;
        }
        ListNode slow = head;
        ListNode fast = head.next;
        
        while(fast!=null && fast.next!=null && slow!=null){
            slow = slow.next;
            fast = fast.next.next;
            if(slow==fast){
                return true;
            }
        }
        return false;
    }

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