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  • 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.
    

    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?

    Accepted
    360,400
    Submissions
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    /**
     * 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;
        }
    }

    /**
     * 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;
        }
    }

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  • 原文地址:https://www.cnblogs.com/Roni-i/p/10430365.html
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