LeetCode:19. Remove Nth Node From End of List-优快云博客

本文链接：https://blog.youkuaiyun.com/Wisimer/article/details/93066949

博客围绕LeetCode第19题展开，题目是给定链表删除倒数第n个结点并返回结果。介绍了使用两个哨兵指针的解题思路，第一个指针先走，第二个指针n步后再走，同时要注意特殊情况。还提到可将问题转化为删除第L - n + 1个结点，需先遍历链表得长度。

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LeetCode:19. Remove Nth Node From End of List

Given a linked list, remove the n-th node from the end of list and return its head.

Example:

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

题目的意思就是给定一个链表，删除倒数第n个结点，返回删除后的链表。

思路

很简单，使用两个哨兵指针，第一个先走，第二个等n步之后再走，这样当第一个走到最后一个结点的时候，第二个指针就处在倒数第n个结点处。当然在实现的时候注意几个特殊情况：一种是倒数第n个，其实就是第一个结点；另一种就是倒数第一个，它的下一个结点为空。

Python 代码实现

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
        
        former = head
        latter = head
        step = 0
        
        if n == 1 and head.next == None:
            return []
        
        while former is not None:
            print(former.val)
            former = former.next
            if step > n:
                latter = latter.next
            step+=1
            
        if latter.next is not None:
            if step > n:
                tmp = latter
                tmp.next = tmp.next.next
                latter.next = tmp.next
            if step ==  n:
                head = latter.next
 
        return head