In this **Leetcode Linked List Cycle II problem solution** we have Given 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 some node in the list can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that the tail's next pointer is connected to. Note that pos is not passed as a parameter. Notice that you should not modify the linked list.

## Problem solution in Python.

class Solution: def hasCycle(self, head: ListNode) -> bool: dict1 = dict() current = head while(current): if(current in dict1): return True dict1[current] = None current = current.next return False

## Problem solution in Java.

public ListNode detectCycle(ListNode head) { ListNode fast = head; ListNode slow = head; do { if (fast == null || fast.next == null) { return null; } slow = slow.next; fast = fast.next.next; } while (fast != slow); slow = head; while (slow != fast) { slow = slow.next; fast = fast.next; } return slow; }

## Problem solution in C++.

class Solution { public: ListNode *detectCycle(ListNode *head) { ListNode *resultNode; ListNode *slow = head; ListNode *fast = head; if(head == nullptr){ resultNode = nullptr; } while(fast !=nullptr && fast->next !=nullptr && slow !=nullptr){ slow = slow->next; fast = fast->next->next; if(slow == fast){ slow = head; while(slow != fast){ slow = slow->next; fast = fast->next; } resultNode = slow; return resultNode; } } resultNode = nullptr; return resultNode; } };

## Problem solution in C.

struct ListNode *detectCycle(struct ListNode *head) { if (head == NULL || head->next == NULL) { return NULL; } struct ListNode* slow = head; struct ListNode* fast = head; 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; } struct ListNode* p1 = head; struct ListNode* p2 = slow; while (p1 != p2) { p1 = p1->next; p2 = p2->next; } return p1; }

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