[LeetCode] 876. Middle of the Linked List

标签：fas   class   NPU   from   out   for   none   code   rom   

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.

这道题是O(N)，想法很简单，快慢指针。注意快慢指针都是从head开始的。每次要判断快指针和快指针下面的node是不是none

# 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:
        if not head: return head
        
        fast, slow = head, head
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
        return slow

[LeetCode] 876. Middle of the Linked List

标签：fas   class   NPU   from   out   for   none   code   rom   

原文地址：https://www.cnblogs.com/codingEskimo/p/12663239.html