給你一個鏈表，刪除鏈表的倒數第?n?個結點，并且返回鏈表的頭結點。

進階：你能嘗試使用一趟掃描實現嗎？

示例 1：

輸入：head = [1,2,3,4,5], n = 2 輸出：[1,2,3,5]

示例 2：

輸入：head = [1], n = 1 輸出：[]

示例 3：

輸入：head = [1,2], n = 1 輸出：[1]

提示：

• 鏈表中結點的數目為?sz
• 1 <= sz <= 30
• 0 <= Node.val <= 100
• 1 <= n <= sz

思路：求出鏈表的長度，刪除倒數第N個節點

# 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: # dummy = ListNode(0, head) # first = head # second = dummy # for i in range(n): # first = first.next# while first: # first = first.next # second = second.next# second.next = second.next.next # return dummy.nextclass Solution:# def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:# def getLength(head: ListNode) -> int:# length = 0# while head:# length += 1# head = head.next# return length# dummy = ListNode(0, head)# length = getLength(head)# cur = dummy# for i in range(1, length - n + 1):# cur = cur.next# cur.next = cur.next.next# return dummy.nextdef removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:def getLength(head:ListNode)->int:length = 0while head:length+=1head = head.nextreturn lengthdummy = ListNode(0,head)length = getLength(head)cur = dummyfor i in range(1,length-n+1):cur = cur.nextcur.next = cur.next.nextreturn dummy.next

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總結

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