[leetcode] 876 intermediate node of linked list

Given a header node as head The non empty single chain table of , Returns the middle node of the linked list .

If there are two intermediate nodes , Then return to the second intermediate node .

Example 1：

Input ： [1,2,3,4,5]
Output ： Nodes in this list 3 ( Serialization form ：[3,4,5])
The returned node value is 3 . ( The evaluation system expresses the serialization of this node as follows [3,4,5]).
Be careful , We returned one ListNode Object of type ans, such ：
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, as well as ans.next.next.next = NULL.

Example 2：

Input ： [1,2,3,4,5,6]
Output ： Nodes in this list 4 ( Serialization form ：[4,5,6])
Because the list has two intermediate nodes , Values, respectively 3 and 4, Let's go back to the second node .

Tips ：

• The number of nodes in a given list is between 1 and 100 Between .

# Get the length of the linked list first , Then traverse again to get the intermediate value 
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution(object):
    def middleNode(self, head):
        my_iter = head
        cnt = 0
        while my_iter:
            my_iter = my_iter.next
            cnt += 1
        my_iter = head
        for i in range(int(cnt/2)):
            my_iter = my_iter.next
        return my_iter

# Press the pointer into the sequence , Then random access takes the intermediate value 
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution(object):
    def middleNode(self, head):
        arr = []
        n = 0
        while head:
            arr.append(head)
            head = head.next
            n+=1
        return arr[int(n/2)]