From 190477400ebf0842343588c4e9e15c5eab412eb4 Mon Sep 17 00:00:00 2001
From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com>
Date: Wed, 3 May 2023 17:22:46 -0500
Subject: [PATCH] =?UTF-8?q?Update=200019.=E5=88=A0=E9=99=A4=E9=93=BE?=
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---
 ...0019.删除链表的倒数第N个节点.md | 28 ++++++++++++-------
 1 file changed, 18 insertions(+), 10 deletions(-)

diff --git a/problems/0019.删除链表的倒数第N个节点.md b/problems/0019.删除链表的倒数第N个节点.md
index a11ff8ba..c6f5bfc7 100644
--- a/problems/0019.删除链表的倒数第N个节点.md
+++ b/problems/0019.删除链表的倒数第N个节点.md
@@ -127,21 +127,29 @@ Python:
 #     def __init__(self, val=0, next=None):
 #         self.val = val
 #         self.next = next
+
 class Solution:
     def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
-        head_dummy = ListNode()
-        head_dummy.next = head
-
-        slow, fast = head_dummy, head_dummy
-        while(n>=0): #fast先往前走n+1步
+        # 创建一个虚拟节点，并将其下一个指针设置为链表的头部
+        dummy_head = ListNode(0, head)
+        
+        # 创建两个指针，慢指针和快指针，并将它们初始化为虚拟节点
+        slow = fast = dummy_head
+        
+        # 快指针比慢指针快 n+1 步
+        for i in range(n+1):
             fast = fast.next
-            n -= 1
-        while(fast!=None):
+        
+        # 移动两个指针，直到快速指针到达链表的末尾
+        while fast:
             slow = slow.next
             fast = fast.next
-        #fast 走到结尾后，slow的下一个节点为倒数第N个节点
-        slow.next = slow.next.next #删除
-        return head_dummy.next
+        
+        # 通过更新第 (n-1) 个节点的 next 指针删除第 n 个节点
+        slow.next = slow.next.next
+        
+        return dummy_head.next
+
 ```
 Go:
 ```Go