# 142. Linked List Cycle II

Given the `head` of a linked list, return _the node where the cycle begins. If there is no cycle, return_ `null`.

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 (**0-indexed**). It is `-1` if there is no cycle. **Note that** `pos` **is not passed as a parameter**.

**Do not modify** the linked list.

&#x20;

**Example 1:**

![](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist.png)

```
Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist_test2.png)

```
Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.
```

**Example 3:**

![](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist_test3.png)

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

&#x20;

**Constraints:**

* The number of the nodes in the list is in the range `[0, 104]`.
* `-105 <= Node.val <= 105`
* `pos` is `-1` or a **valid index** in the linked-list.

&#x20;

**Follow up:** Can you solve it using `O(1)` (i.e. constant) memory?

解题思路:

快慢两个指针,

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