Middle of the Linked List Python

Given a non-empty, singly linked list with head node head, return a middle node of linked list.
If there are two middle nodes, return the second middle node.

Example 1:
Input: [1,2,3,4,5]
Output: Node 3 from this list (Serialization: [3,4,5])
The returned node has value 3.  (The judge's serialization of this node is [3,4,5]).
Note that we returned a ListNode object ans, such that:
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.
Example 2:
Input: [1,2,3,4,5,6]
Output: Node 4 from this list (Serialization: [4,5,6])
Since the list has two middle nodes with values 3 and 4, we return the second one.

Note:
  • The number of nodes in the given list will be between 1 and 100.

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def middleNode(self, head: ListNode) -> ListNode:
        pA = head
        if head is None:
            return head
        pA = head
        lA = 0
        while pA:
            lA+=1
            pA =pA.next
        pB = head
        lA = lA//2
        while lA != 0:
            pB = pB.next
            lA -= 1
        return pB if lA % 2 ==0 else pB.next
            
Time Complexity : O(n)
Space Complexity : O(1)


Fast Pointer and Slow Pointer :


# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def middleNode(self, head: ListNode) -> ListNode:
        slow = fast = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        return slow
    
Time Complexity : O(n)
Space Complexity : O(1)           
            
        
        

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