[IMP] Linked-List-1

linked-list-cycle-ii.py

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+'''
+Time Complexity : O(m + n)
+Space Complexity : O(1) 
+Did this code successfully run on Leetcode : Yes
+Any problem you faced while coding this :  No
+'''
+# Definition for singly-linked list.
+class ListNode:
+    def __init__(self, x):
+        self.val = x
+        self.next = None
+
+class Solution:
+    def detectCycle(self, head):
+        slow = head
+        fast = head
+
+        while fast is not None and fast.next is not None:
+            slow = slow.next
+            fast = fast.next.next
+
+            if slow == fast:
+                fast = head
+                while slow != fast:
+                    slow = slow.next
+                    fast = fast.next
+                return slow
+

remove-nth-node-from-end-of-list.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
+'''
+# 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, n):
+        if head is None:
+            return head
+        dummy = ListNode(0, head)
+        slow = dummy
+        fast = dummy
+        count = 0
+        while count <= n:
+            fast = fast.next
+            count += 1
+
+        while fast is not None:
+            slow = slow.next
+            fast = fast.next
+
+        slow.next = slow.next.next
+
+        return dummy.next
+

reverse-linked-list.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
+'''
+# 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):
+        prev = None
+        curr = head
+        while curr:
+            fast = curr.next
+            curr.next = prev
+            prev = curr
+            curr = fast
+
+        return prev