feat(work): add solution for LeetCode 141 Linked List Cycle

work/1/Easy/Hash Table/141.linked-list-cycle.py

@@ -0,0 +1,73 @@
+"""
+141. Linked List Cycle
+Difficulty: Easy
+https://leetcode.com/problems/linked-list-cycle/
+
+──────────────────────────────────────────────────
+
+Given head, the head of a linked list, determine if the linked list
+has a cycle in it.
+
+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. Note that pos is not passed as a
+parameter.
+
+Return true if there is a cycle in the linked list. Otherwise, return
+false.
+
+
+
+Example 1:
+
+Input: head = [3,2,0,-4], pos = 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], pos = 0
+Output: true
+Explanation: There is a cycle in the linked list, where the tail
+connects to the 0th node.
+
+Example 3:
+
+Input: head = [1], pos = -1
+Output: false
+Explanation: There is no cycle in the linked list.
+
+
+
+Constraints:
+
+        • The number of the nodes in the list is in the range [0, 10^4].
+
+        • -10^5 <= Node.val <= 10^5
+
+        • pos is -1 or a valid index in the linked-list.
+
+
+
+Follow up: Can you solve it using O(1) (i.e. constant) memory?
+"""
+
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+
+class Solution:
+    def hasCycle(self, head: Optional[ListNode]) -> bool:
+        fast = slow = head
+
+        while fast and fast.next:
+            fast = fast.next.next
+            slow = slow.next
+            if fast is slow:
+                return True
+        return False