描述
给定一个链表,删除链表的倒数第 n 个节点,并且返回链表的头结点。
例子
思路 删除结点时,要有头结点
- 删除结点要先找到需删除结点的前一个结点cur, 然后cur.next=cur.next.next。为了解决删除第一个结点时的问题,可以先给链表添加一个头结点。
- 我自己的 遍历两次链表
先给链表构造一个头结点,遍历链表,计算一下链表有几个结点,然后,再次遍历链表,到第num-n个结点,cur.next=cur.next.next。【倒数第n个的前一个结点为正数的num-n个结点】
- 更优 遍历一次链表
- 先给链表构造一个头结点,在设置快慢指针,快指针比慢指针先走(n+1)步,当快指针走到Null结点时,满指针走到倒数第(n+1)个结点
答案 - python
*我自己的*
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
cur=head
num=0
while cur:
cur=cur.next
num+=1
if num-n==0: return head.next
count=1
cur=head
while count!=num-n:
cur=cur.next
count+=1
cur.next = cur.next.next
return head
*优化*
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
newHead=ListNode(0)
newHead.next = head
slow=newHead
fast=newHead
for i in range(n+1):
fast=fast.next
while fast:
fast=fast.next
slow=slow.next
slow.next = slow.next.next
return newHead.next
- c++
*我自己*
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* newHead = new ListNode(0);
newHead->next = head;
ListNode* cur=newHead;
int num=0;
while (cur)
{
num+=1;
cur = cur->next;
}
int count = 1;
cur=newHead;
while (count!=num-n)
{
count+=1;
cur=cur->next;
}
cur->next = cur->next->next;
return newHead->next;
}
};
*优化*
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* newHead = new ListNode(0);
newHead->next = head;
ListNode *slow=newHead,*fast=newHead;
for (int i=0; i<n+1; i++)
fast = fast->next;
while (fast)
{
fast=fast->next;
slow=slow->next;
}
slow->next = slow->next->next;
return newHead->next;
}
};