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)
设置快慢两个指针,若有环,两个指针定会相遇
public boolean hasCycle(ListNode head) { ListNode oneStep = head; ListNode twoStep =head; while(null!=oneStep&&null!=twoStep&&null!=twoStep.next){ oneStep=oneStep.next; twoStep=twoStep.next.next; if (oneStep==twoStep){ return true; } } return false; }
进阶题
查询有环链表的环入口 LeetCode142 https://www.cnblogs.com/zhacai/p/10561152.html