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.

->要找出此linkedlist中是否有環

<解題>

1. 用快和慢兩指針，快的一次走兩步、慢的一次走一步，如果兩者最後能後相遇，代表當中有環
2. 可以先在一開始確認(head == null)的話，就直接return false

public class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null) {
            return false;
        }

        ListNode slow = head;
        ListNode fast = head.next;

        while (slow != fast && fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }

        return slow == fast;
    }
}

Time Complexity: O(N) Space Complexity: O(1)