9/10 每日一題(鏈結串列 回傳中間以後的鏈結)

 # Definition for singly-linked list.

class ListNode:

    def __init__(self, val=0, next=None):

        self.val = val

        self.next = next

       

class Solution:  

    def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]:

        n=0

        p=head

        while p:

            n+=1

            p=p.next

       

        p=head

        mid=n//2

        n=0

        while p:

            if n == mid:

                head=p

        #來到中間點,回傳剩下的鏈結串列

                return head

            n+=1

            p=p.next

--------------------------------使用雙指針

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]:
        fast = head
        slow = head

        while(fast and fast.next): #避免fast在None,沒有next屬性

            fast = fast.next.next
            slow = slow.next
        return slow

-----------------------------------------------

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]:
        temp=head
        c=0
        while(temp!=None):
            c=c+1
            temp=temp.next 
        if(c%2!=0):
            pos=c//2 
            temp=head
            i=1
            while(i<=pos):
                temp=temp.next 
                i=i+1
            return temp
        else:
            pos=c//2
            temp=head
            i=1
            while(i<=pos):
                temp=temp.next
                i=i+1
            return temp