#19. 删除链表的倒数第 N 个结点

Posted 2021-05-20 王爷爱吃秋刀鱼

题目：

 

 

思路：一道经典的快慢指针问题，快指针先走n步，然后一起走，快指针走到最后一个节点，慢指针就到了要被删除的节点的前一个节点。

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        node = ListNode(0)
        node.next = head #定义头节点
        fast = node
        slow = node
        for i in range(n):
            fast = fast.next #快节点先走n步
        while(fast.next!=None):#直到快节点走到最后一个，慢节点直接指向她的后继节点的后继节点
            fast = fast.next
            slow = slow.next
        slow.next = slow.next.next
        return node.next