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:
1 and 100.Source
# Definition for singly-linked list.
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
def middle_node(head):
"""Naive solution: Counting the number of nodes"""
node = head
counter = 0
while node.next:
counter += 1
node = node.next
counter = (counter+1) // 2 # counter+1 to account for the last node
node = head
while counter > 0:
counter -= 1
node = node.next
return node
def middle_node(head):
"""Using 2 pointers."""
if head is None:
return head
slow, fast = head, head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
return slow