linked-list-1

LinkedListCycle2.py

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+# Time Complexity : O(n)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach : Use two pointers moving at different speeds to check if there’s a cycle.
+# If they meet, reset one to the head and walk both one step at a time.
+# They meet again at the node where the cycle starts.
+
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        if head is None:
+            return None
+
+        slow, fast = head, head
+        flag = False
+
+        while fast.next is not None and fast.next.next is not None:
+            slow = slow.next
+            fast = fast.next.next
+
+            if slow == fast:
+                flag = True
+                break
+
+        if not flag:
+            return None
+
+        slow = head
+        while slow != fast:
+            slow = slow.next
+            fast = fast.next
+
+        return slow
+
+
+

RemoveNthNode.py

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+# Time Complexity : O(n)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach : Move fast pointer n + 1 steps ahead so it leads the slow by n nodes.
+# Then increment both together until fast hits the end — now slow is right before the target.
+# Skip the node to be removed and clean up the reference.
+
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
+        dummy = ListNode(-1)
+        dummy.next = head
+        slow, fast = dummy, dummy
+
+        for i in range(n+1):
+            fast = fast.next
+
+        while fast is not None:
+            slow = slow.next
+            fast = fast.next
+
+        temp = slow.next
+        slow.next = slow.next.next
+        temp.next = None
+
+        return dummy.next
+

ReverseLinkedList.py

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+# Time Complexity : O(n)
+# Space Complexity : O(1)
+# Did this code successfully run on Leetcode : Yes
+# Any problem you faced while coding this : No
+# Approach : Flip the direction of the pointer to the previous node.
+
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        if head is None:
+            return None
+
+        prev, curr = None, head
+
+        while curr:
+            temp = curr.next
+            curr.next = prev
+            prev = curr
+            curr = temp
+
+        return prev
+
+
+