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‎Problemset/0142-Linked List Cycle II/README.md

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# 142. Linked List Cycle II
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Given the `head` of a linked list, return *the node where the cycle begins. If there is no cycle, return* `null`.
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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**.
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**Do not modify** the linked list.
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#### Example 1:
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![](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist.png)
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<pre>
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<strong>Input:</strong> head = [3,2,0,-4], pos = 1
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<strong>Output:</strong> tail connects to node index 1
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<strong>Explanation:</strong> There is a cycle in the linked list, where tail connects to the second node.
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</pre>
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#### Example 2:
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![](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist_test2.png)
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<pre>
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<strong>Input:</strong> head = [1,2], pos = 0
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<strong>Output:</strong> tail connects to node index 0
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<strong>Explanation:</strong> There is a cycle in the linked list, where tail connects to the first node.
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</pre>
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#### Example 3:
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![](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist_test3.png)
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<pre>
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<strong>Input:</strong> head = [1], pos = -1
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<strong>Output:</strong> no cycle
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<strong>Explanation:</strong> There is no cycle in the linked list.
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</pre>
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#### Constraints:
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* The number of the nodes in the list is in the range <code>[0, 10<sup>4</sup>]</code>.
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* <code>-10<sup>5</sup> <= Node.val <= 10<sup>5</sup></code>
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* `pos` is `-1` or a **valid index** in the linked-list.
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**Follow up:** Can you solve it using `O(1)` (i.e. constant) memory?
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## Solutions (Python)
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### 1. Solution
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```Python
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, x):
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# self.val = x
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# self.next = None
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class Solution:
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def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
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if head is None or head.next is None:
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return None
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slow = head.next
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fast = head.next.next
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while slow != fast:
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if fast is None or fast.next is None:
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return None
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slow = slow.next
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fast = fast.next.next
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while head != slow:
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head = head.next
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slow = slow.next
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return head
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```

‎Problemset/0142-Linked List Cycle II/README_CN.md

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# 142. 环形链表 II
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给定一个链表的头节点 `head` ，返回链表开始入环的第一个节点。 *如果链表无环，则返回 `null`*。
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如果链表中有某个节点，可以通过连续跟踪 `next` 指针再次到达，则链表中存在环。 为了表示给定链表中的环，评测系统内部使用整数 `pos` 来表示链表尾连接到链表中的位置（**索引从 0 开始**）。如果 `pos` 是 `-1`，则在该链表中没有环。**注意：**`pos` **不作为参数进行传递**，仅仅是为了标识链表的实际情况。
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**不允许修改** 链表。
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#### 示例 1:
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![](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist.png)
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<pre>
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<strong>输入:</strong> head = [3,2,0,-4], pos = 1
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<strong>输出:</strong> 返回索引为 1 的链表节点
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<strong>解释:</strong> 链表中有一个环，其尾部连接到第二个节点。
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</pre>
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#### 示例 2:
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![](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist_test2.png)
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<pre>
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<strong>输入:</strong> head = [1,2], pos = 0
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<strong>输出:</strong> 返回索引为 0 的链表节点
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<strong>解释:</strong> 链表中有一个环，其尾部连接到第一个节点。
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</pre>
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#### 示例 3:
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![](https://assets.leetcode.com/uploads/2018/12/07/circularlinkedlist_test3.png)
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<pre>
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<strong>输入:</strong> head = [1], pos = -1
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<strong>输出:</strong> 返回 null
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<strong>解释:</strong> 链表中没有环。
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</pre>
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#### 提示:
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* 链表中节点的数目范围在范围 <code>[0, 10<sup>4</sup>]</code> 内
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* <code>-10<sup>5</sup> <= Node.val <= 10<sup>5</sup></code>
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* `pos` 的值为 `-1` 或者链表中的一个有效索引
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**进阶：**你是否可以使用 `O(1)` 空间解决此题？
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## 题解 (Python)
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### 1. 题解
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```Python
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, x):
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# self.val = x
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# self.next = None
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class Solution:
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def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
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if head is None or head.next is None:
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return None
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slow = head.next
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fast = head.next.next
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while slow != fast:
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if fast is None or fast.next is None:
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return None
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slow = slow.next
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fast = fast.next.next
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while head != slow:
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head = head.next
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slow = slow.next
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return head
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```

‎Problemset/0142-Linked List Cycle II/Solution.py

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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, x):
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# self.val = x
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# self.next = None
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class Solution:
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def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
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if head is None or head.next is None:
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return None
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slow = head.next
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fast = head.next.next
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while slow != fast:
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if fast is None or fast.next is None:
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return None
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slow = slow.next
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fast = fast.next.next
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while head != slow:
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head = head.next
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slow = slow.next
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return head

‎README.md

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[138][138l] |[Copy List with Random Pointer][138] |![py]
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[139][139l] |[Word Break][139] |![py]&nbsp;&nbsp;![rb]
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[141][141l] |[Linked List Cycle][141] |![py]
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[142][142l] |[Linked List Cycle II][142] |![py]
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[144][144l] |[Binary Tree Preorder Traversal][144] |![py]
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[145][145l] |[Binary Tree Postorder Traversal][145] |![py]
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[147][147l] |[Insertion Sort List][147] |![py]
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[138]:Problemset/0138-Copy%20List%20with%20Random%20Pointer/README.md#138-copy-list-with-random-pointer
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[139]:Problemset/0139-Word%20Break/README.md#139-word-break
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[141]:Problemset/0141-Linked%20List%20Cycle/README.md#141-linked-list-cycle
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[142]:Problemset/0142-Linked%20List%20Cycle%20II/README.md#142-linked-list-cycle-ii
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[144]:Problemset/0144-Binary%20Tree%20Preorder%20Traversal/README.md#144-binary-tree-preorder-traversal
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[145]:Problemset/0145-Binary%20Tree%20Postorder%20Traversal/README.md#145-binary-tree-postorder-traversal
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[147]:Problemset/0147-Insertion%20Sort%20List/README.md#147-insertion-sort-list

‎README_CN.md

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[138][138l] |[复制带随机指针的链表][138] |![py]
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[139][139l] |[单词拆分][139] |![py]&nbsp;&nbsp;![rb]
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[141][141l] |[环形链表][141] |![py]
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[142][142l] |[环形链表 II][142] |![py]
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[144][144l] |[二叉树的前序遍历][144] |![py]
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[145][145l] |[二叉树的后序遍历][145] |![py]
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[147][147l] |[对链表进行插入排序][147] |![py]
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[138]:Problemset/0138-Copy%20List%20with%20Random%20Pointer/README_CN.md#138-复制带随机指针的链表
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[139]:Problemset/0139-Word%20Break/README_CN.md#139-单词拆分
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[141]:Problemset/0141-Linked%20List%20Cycle/README_CN.md#141-环形链表
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[142]:Problemset/0142-Linked%20List%20Cycle%20II/README_CN.md#142-环形链表-ii
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[144]:Problemset/0144-Binary%20Tree%20Preorder%20Traversal/README_CN.md#144-二叉树的前序遍历
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[145]:Problemset/0145-Binary%20Tree%20Postorder%20Traversal/README_CN.md#145-二叉树的后序遍历
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[147]:Problemset/0147-Insertion%20Sort%20List/README_CN.md#147-对链表进行插入排序

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