[LeetCode] 876. Middle of the Linked List_Easy tag: Linked List
2019-05-02 10:12 Johnson_强生仔仔 阅读(299) 评论(0) 编辑 收藏 举报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
and100
.
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