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 and 100.
Definition for singly-linked list. class ListNode: def __init__(self, x): self.val = x self.next = None class Solution: def middleNode(self, head): a=head b=head b=b.next while(b!=None): a=a.next b=b.next if(b!=None): b=b.next return a