Add solution for cycle linked list

neetcode/linked_list/linked_list_cycles/README.md

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+Linked List Cycle Detection
+Given the beginning of a linked list head, return true if there is a cycle in the linked list. Otherwise, return false.
+
+There is a cycle in a linked list if at least one node in the list that can be visited again by following the next pointer.
+
+Internally, index determines the index of the beginning of the cycle, if it exists. The tail node of the list will set it's next pointer to the index-th node. If index = -1, then the tail node points to null and no cycle exists.
+
+Note: index is not given to you as a parameter.
+
+Example 1:
+
+
+
+Input: head = [1,2,3,4], index = 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], index = -1
+
+Output: false
+Constraints:
+
+1 <= Length of the list <= 1000.
+-1000 <= Node.val <= 1000
+index is -1 or a valid index in the linked list.
+
+# Solution
+## Использовать хэшмап
+
+nodes = defaultdict()

neetcode/linked_list/linked_list_cycles/solution.py

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+from collections import defaultdict
+from typing import Optional
+
+
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+
+class Solution:
+    def hasCycle(self, head: Optional[ListNode]) -> bool:
+        nodes: set[ListNode] = set()
+        cur_node = head
+        while cur_node:
+            lb = len(nodes)
+            nodes.add(cur_node)
+            cur_node = cur_node.next
+            la = len(nodes)
+            if lb == la:
+                return True
+        return False
+
+
+if __name__ == "__main__":
+    s = Solution()
+    e3 = ListNode(3, None)
+    e2 = ListNode(2, e3)
+    e1 = ListNode(1, e2)
+    e0 = ListNode(0, e1)
+    # k = [e0, e1, e2, e3]
+    nb = s.hasCycle(e0)
+    print(nb)