【双指针-中等】19. 删除链表的倒数第 N 个结点

【题目】

【代码】
【方法1】在head之前设置提供特殊节点mute用于保存head，mute的next指针指向head节点。

class Solution:
    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
        mute=ListNode()
        mute.val=-1
        mute.next=head
        fast=head
        slow=head
        pre=mute
        while n:
            n-=1
            fast=fast.next
        while fast:
            pre=slow
            slow=slow.next
            fast=fast.next
        pre.next=slow.next
        return mute.next

【方法2】
没有头部多余的节点

class Solution:
    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
        origin=head
        del_p=head
        pre=head
        while n and head.next:
            n-=1
            head=head.next
        if n!=0:
            return origin.next
        while head:
            head=head.next
            pre=del_p
            del_p=del_p.next
        pre.next=del_p.next
        return origin