141. Linked List Cycle

Difficulty: Easy
17 / 17 test cases passed.
Runtime: 36 ms
Memory Usage: 17.4 MB

Easy/141.LinkedListCycle.py

@@ -0,0 +1,59 @@
+"""
+    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:
+    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:
+    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:
+    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?
+"""
+#Difficulty: Easy
+#17 / 17 test cases passed.
+#Runtime: 36 ms
+#Memory Usage: 17.4 MB
+
+#Runtime: 36 ms, faster than 99.38% of Python3 online submissions for Linked List Cycle.
+#Memory Usage: 17.4 MB, less than 99.22% of Python3 online submissions for Linked List Cycle.
+
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def hasCycle(self, head: ListNode) -> bool:
+        s = set()
+        while head:
+            if head in s:
+                return True
+            s.add(head)
+            head = head.next
+        return False