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
fast pointer to find the middle number. slow pointer to get to the middle node.
# 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 head == None: return head count = 0 fast = head while fast != None: fast = fast.next count += 1 mid = count // 2 + 1 slow = head for i in range(1, mid): slow = slow.next return slow
- Time complexity: O(n).
- Space complexity: O(1).