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 next
pointer. Internally, pos
is used to denote the index of the node that tail’s next
pointer is connected to. Note that pos
is not passed as a parameter.
Return true
if there is a cycle in the linked list. Otherwise, return false
.
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:
[0, 104]
.-105 <= Node.val <= 105
pos
is -1
or a valid index in the linked-list.Follow up: Can you solve it using O(1)
(i.e. constant) memory?
using LeetCodeNet.Com_github_leetcode;
/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public bool HasCycle(ListNode head) {
if (head == null) {
return false;
}
ListNode fast = head.next;
ListNode slow = head;
while (fast != null && fast.next != null) {
if (fast == slow) {
return true;
}
fast = fast.next.next;
slow = slow.next;
}
return false;
}
}