Class Solution
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- g0101_0200.s0141_linked_list_cycle.Solution
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public class Solution extends Object
141 - Linked List Cycle.Easy
Given
head, the head of a linked list, determine if the linked list has a cycle in it.There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the
nextpointer. Internally,posis used to denote the index of the node that tail’snextpointer is connected to. Note thatposis not passed as a parameter.Return
trueif there is a cycle in the linked list. Otherwise, returnfalse.Example 1:

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

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

Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.
Constraints:
- The number of the nodes in the list is in the range
[0, 104]. -105 <= Node.val <= 105posis-1or a valid index in the linked-list.
Follow up: Can you solve it using
O(1)(i.e. constant) memory? - The number of the nodes in the list is in the range
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Constructor Summary
Constructors Constructor Description Solution()
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Method Detail
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hasCycle
public boolean hasCycle(ListNode head)
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