chore: add daily leetcode post 剑指 Offer II 021. 删除链表的倒数第 n 个结点_translated

Author: longsizhuo

app/docs/CommunityShare/Leetcode/剑指 Offer II 021. 删除链表的倒数第 n 个结点_translated.md

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+---
+title: Sword finger Offer II 021. Delete the countdown of the linked list n Node.md
+date: '2024.01.01 0:00'
+tags:
+  - - Python
+  - - answer
+abbrlink: 3ed2f01c
+---
+
+# topic：
+
+[Sword finger Offer II 021. Delete the countdown of the linked list n Node.md](https://leetcode.cn/problems/SLwz0R/description/)
+
+# Thought：
+Double pointer（Sliding window algorithm）。
+In this method，We first created a virtual head node dummy，And point it to the original head point head。
+Then we use two pointers fast and slow，Will fast Poor movement move forward n step。
+Next，We move at the same time fast and slow pointer，until fast pointer到达链表的末尾。
+at this time，slow pointer指向倒数第 n+1 Node，我们Will其 next pointer指向 slow.next.next，So as to delete the countdown n Node。
+
+at last，We return virtual head nodes next pointer，It points to delete the countdown n Node后的链表的Head node。
+
+At the beginning, according toCLinked
+# Code：
+
+```python   Sliding window algorithm
+class Solution:
+    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
+        # Create a virtual head node
+        dummy = ListNode(0)
+        # Will虚拟Head node的 next Pointing to the original head point
+        dummy.next = head
+        # 定义快慢pointer，并Will快Poor movement move forward n step
+        fast = slow = dummy
+        for i in range(n):
+            fast = fast.next
+        # 同时移动快慢pointer，until快pointer到达链表末尾
+        while fast and fast.next:
+            fast = fast.next
+            slow = slow.next
+
+        # Will慢pointer的 next pointer指向慢pointer的下一Node的下一Node，So as to delete the countdown n Node
+        slow.next = slow.next.next
+        return dummy.next
+```
+
+```python Pure linked list
+class Solution:
+    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
+
+        # Calculation length
+        def get_list_length(head):
+            # If the linked list is empty，Length0
+            if not head:
+                return 0
+
+            # Links in traversal，Count
+            length = 0
+            current = head
+            while current:
+                length += 1
+                current = current.next
+
+            return length
+
+        # Find the deleted node
+        def delete(node, count):
+            if count == n + 1 or n == length:
+                node.next = node.next.next
+                return
+            if node.next:
+                delete(node.next, count - 1)
+
+        length = get_list_length(head)
+        delete(head, length)
+        return head
+
+
+def list_to_linked_list(lst):
+    if not lst:
+        return None
+
+    # Head node
+    head = ListNode(lst[0])
+    current = head
+
+    # Elements in the list of traversal，Will其转换为链表节点
+    for i in range(1, len(lst)):
+        current.next = ListNode(lst[i])
+        current = current.next
+
+    return head
+```