[LeetCode] 876. Middle of the Linked List_Easy tag: Linked List ** slow, fast pointers

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.

 

Note:

• The number of nodes in the given list will be between 1 and 100.

 

思路直接求出来linked list的长度, 然后再走一半, 返回head即可.

Improve, 可以用slow 和fast 去走, 当fast走到头, 那么slow就在中间了.

 

Code    T: O(n)    S: O(1)

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def middleNode(self, head):
        def helper(head):
            count = 0
            while head:
                count += 1
                head = head.next
            return count
        
        for i in range(helper(head)//2):
            head = head.next
        return head