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# Linked List Cycle II LeetCode Solution

## Problem – 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.

Example 1:

``````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:

``````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:

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

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.

### Linked List Cycle II LeetCode Solution in Python

``````class Solution(object):
while fast and fast.next:
slow, fast = slow.next, fast.next.next
if slow == fast: break
else: return None  # if not (fast and fast.next): return None
``````

### Linked List Cycle II LeetCode Solution in C++

``````class Solution {
public:
while (fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
if (slow == fast) break;
}
if (!(fast && fast->next)) return NULL;
slow = slow->next;
}
}
};
``````

### Linked List Cycle II LeetCode Solution in Java

``````public class Solution {
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if (slow == fast) break;
}
if (fast == null || fast.next == null) return null;
slow = slow.next;
}
}
}
``````
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