[Leetcode] linked list cycle

Linked List cycle

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

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

Algorithm

Thought 1:

Fast and slow pointers. When two pointers meet each other, it indicates that there is a ring. Otherwise, there is no ring. It may need to be traversed multiple times. The space complexity is O (1), and the time complexity is O (n)

 1 public class Solution { 2     public boolean hasCycle(ListNode head) { 3         if(head == null) return false; 4         ListNode fast = head; 5         ListNode slow = head; 6         while(true){ 7             if(fast.next == null || fast.next.next == null) return false; 8             fast = fast.next.next; 9             slow = slow.next;10             if(fast == slow) return true;11         }12     }13 }

 

Idea 2:

Hash table. When a node has two prefixes, the node is the starting point of the ring and only needs to be traversed once. The space complexity is O (n), and the time complexity is O (n)

 1 public class Solution { 2     public boolean hasCycle(ListNode head) { 3         if(head == null) return false; 4         ListNode tem = new ListNode(0); 5         tem.next = head; 6         Map<ListNode,ListNode> hash = new HashMap<ListNode,ListNode>(); 7         while(true){ 8             if(tem.next == null) return false; 9             if(hash.get(tem.next) != null) return true;10             hash.put(tem.next,tem);11             tem = tem.next;12         }13     }14 }

Idea 2 is the solution of linked list cycle II.

 

The listnode structure is as follows:

1 public class ListNode {2     int val;3     ListNode next;4     ListNode(int x){5         val = x;6         next = null;7     }8 }

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