Given the head of a singly linked list, return the middle node of the linked list.
If there are two middle nodes, return the second middle node.
Example 1:
Input: head = [1,2,3,4,5] Output: [3,4,5] Explanation: The middle node of the list is node 3.
Example 2:
Input: head = [1,2,3,4,5,6] Output: [4,5,6] Explanation: Since the list has two middle nodes with values 3 and 4, we return the second one.
Constraints:
The number of nodes in the list is in the range [1, 100]. 1 <= Node.val <= 100
Solution
Approach 1: Two Passes
- Find the length of the linked list by traversing the list.
- Find the middle node by traversing the list again to the middle.
- Return the middle node
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
def middle_node(head: ListNode) -> ListNode:
length = 0
current = head
while current:
length += 1
current = current.next
middle = length // 2
current = head
for _ in range(middle):
current = current.next
return current
Time complexity: O(n) Space complexity: O(1)
https://youtu.be/xcsDK3mziCM?embedable=true
Approach 2: One Pass
- Use two pointers, one fast and one slow.
- Move the fast pointer two nodes at a time and the slow pointer one node at a time.
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
def middle_node(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)