标签:head 141 list List fast next 链表 linked cycle
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.
判断链表是否有环:使用快慢指针去判断,如果有环,fast快指针先进入环,slow慢指针后进入,经过一定步的操作,快慢指针在环上相遇。
方法一:(C++)
1 class Solution { 2 public: 3 bool hasCycle(ListNode *head) { 4 ListNode *fast=head,*slow=head; 5 while(fast&&fast->next){ 6 slow=slow->next; 7 fast=fast->next->next; 8 if(slow==fast) 9 return true; 10 } 11 return false; 12 } 13 };
方法二:(Java)
1 public class Solution { 2 public boolean hasCycle(ListNode head) { 3 Set<ListNode> s=new HashSet(); 4 while(head!=null){ 5 if(s.contains(head)) 6 return true; 7 else 8 s.add(head); 9 head=head.next; 10 } 11 return false; 12 } 13 }
Java集合中含有contains()方法,判断其是否已经含有此节点,如果没有,则使用add()添加进去
标签:head,141,list,List,fast,next,链表,linked,cycle 来源: https://www.cnblogs.com/hhhhan1025/p/10582829.html
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