141. Linked List Cycle【Easy】【判断链表是否存在环】

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.

141. Linked List Cycle【Easy】【判断链表是否存在环】

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.

141. Linked List Cycle【Easy】【判断链表是否存在环】

Example 3:

Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

141. Linked List Cycle【Easy】【判断链表是否存在环】

Follow up:

Can you solve it using O(1) (i.e. constant) memory?

Accepted
360,400
Submissions
1,010,955
 
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public boolean hasCycle(ListNode head) {
if(head == null) return false;
ListNode slow = head;
ListNode fast = head.next;
while(slow!=null && fast!= null && fast.next!=null) {
slow = slow.next;
fast = fast.next.next;
if(slow == fast) return true;
}
return false;
}
}

141. Linked List Cycle【Easy】【判断链表是否存在环】

/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public boolean hasCycle(ListNode head) {
ListNode slow = head, fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if (slow == fast) return true;
}
return false;
}
}

141. Linked List Cycle【Easy】【判断链表是否存在环】

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