[LeetCode] 876. Middle of the Linked List_Easy tag: Linked List

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.

这个题目暴力做法就是将list走一遍，然后记录个数为n，接着再走n/2步即可。

但是这里给出用fast/slow pointers的做法，也就是用两个指针，fast一次走两步，slow一次走一步。如果fast.next is None, return slow; if fast.next.next is None, return slow.next(可以自己用一个小例子来判断就好)

Update on 09/20/2023, cleaner code 

Code

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def middleList(self, head):
        # if not head: return head   #dont need because length belongs [1, 100]
        slow, fast = head, head
        while fast:
            if fast.next is None: # 如果length是偶数，slow要在前面的话就将下一个if和这个if or起来。
                return slow
            if fast.next.next is None:
                return slow.next
            slow = slow.next
            fast = fast.next.next
    

 

Code

class Solution:
    def middleList(self, head):
        dummy = ListNode(0)
        dummy.next = head
        fast, slow = dummy, dummy
        while slow:
            if not fast.next or not fast.next.next:
                return slow.next
            slow = slow.next
            fast = fast.next.next

 

Code

class Solution:
    def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]:
        fast, slow = head, head
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
        return slow