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

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

 

Follow up:

Can you solve it using O(1) (i.e. constant) memory?

空间复杂度O(n)

public boolean hasCycle(ListNode head) {//链表 
        Set<ListNode> nodeSet = new HashSet<ListNode>();
        while(null!= head){
            if(nodeSet.contains(head)){
                return true;
            }
            else{
                nodeSet.add(head);
                head=head.next;
            }
        }
        return false;
    }　　

空间复杂度O(1)

public boolean hasCycle(ListNode head) {//链表 mytip
        ListNode oneStep = head;
        ListNode twoStep  =head;
        while(null!=oneStep){
            oneStep=oneStep.next;
            if(null==twoStep.next||null==twoStep.next.next){
                return false;
            }
            twoStep=twoStep.next.next;
            if (oneStep==twoStep){
                return true;
            }
        }
        return false;
    }