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.
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.
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.
- 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:
p1 = head
p2 = head
while p2 and p2.next:
p1 = p1.next
p2 = p2.next.next
return p1
# 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:
list_len = 0
p = head
while p:
list_len += 1
p = p.next
for i in range(list_len // 2):
head = head.next
return head