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
and100
.
# 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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