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. Two Pointers
# Time: O(n) # Space: O(1) # 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: if not head or (head and not head.next): return head fast, slow = head, head while fast: if not fast.next: break fast = fast.next.next slow = slow.next return slow # Runtime: 32 ms, faster than 35.49% of Python3 # Memory Usage: 13.6 MB, less than 7.14% of Python3