[LeetCode]#876. Middle of the Linked List

Environment: Python 3.8

Key technique: List node

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.

Example 1:

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.

Example 2:

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.

Analysis:

1. I am not so good at linked-list structure, so I add linked-list as below.

class ListNode:
     def __init__(self, val=0, next=None):
         self.val = val
         self.next = next
input=[1,2,3,4,5]
head=ListNode(1)
head.next=ListNode(2)
head.next.next=ListNode(3)
head.next.next.next=ListNode(4)
head.next.next.next.next=ListNode(5)

2. And I see below data.

3. Find all .next data and find middle number.

Solution:

class Solution():
    def middleNode(self, head):
        A = [head]
        while A[-1].next:
            A.append(A[-1].next)
        return A[int(len(A) / 2)]

Submissions:

Reference:

https://leetcode.com/problems/middle-of-the-linked-list/solution/