[Python-leetcode876- speed of the intermediate node linked list pointer]

Problem Description:

Given a non-empty list with a single head of the first node, an intermediate node list returned.

If there are two intermediate nodes, the second intermediate node is returned.

Example 1:

Input: [1,2,3,4,5]
Output: node 3 (SEQ form: [3,4,5]) in the list of
returned node is 3. (This evaluation system is expression of the sequence node [3,4,5]).
Note that, we return an object of type ListNode ANS, so 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: this list node 4 (SEQ form: [4,5,6])
Since the list has two intermediate nodes, values of 3 and 4, we return to the second node.
 

prompt:

Given the list of nodes is between 1 and 100.

 

Core: a slow pointer each time step, a quick pointer each take two steps when the pointer quickly went to the tail, then slow pointer node is an intermediate 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:
        slow = head
        fast = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        return slow

result: