0019.删除链表的倒数第N个节点
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# 19.删除链表的倒数第N个节点
[力扣题目链接](https://leetcode.cn/problems/remove-nth-node-from-end-of-list/)
给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。
进阶：你能尝试使用一趟扫描实现吗？
示例 1：
输入：head = [1,2,3,4,5], n = 2
输出：[1,2,3,5]
示例 2：
输入：head = [1], n = 1
输出：[]
示例 3：
输入：head = [1,2], n = 1
输出：[1]
## 算法公开课
**[《代码随想录》算法视频公开课](https://programmercarl.com/other/gongkaike.html)：：[链表遍历学清楚！ | LeetCode：19.删除链表倒数第N个节点](https://www.bilibili.com/video/BV1vW4y1U7Gf)，相信结合视频再看本篇题解，更有助于大家对链表的理解。**
## 思路
双指针的经典应用，如果要删除倒数第n个节点，让fast移动n步，然后让fast和slow同时移动，直到fast指向链表末尾。删掉slow所指向的节点就可以了。
思路是这样的，但要注意一些细节。
分为如下几步：
* 首先这里我推荐大家使用虚拟头结点，这样方便处理删除实际头结点的逻辑，如果虚拟头结点不清楚，可以看这篇： [链表：听说用虚拟头节点会方便很多？](https://programmercarl.com/0203.移除链表元素.html)
* 定义fast指针和slow指针，初始值为虚拟头结点，如图：
* fast首先走n + 1步 ，为什么是n+1呢，因为只有这样同时移动的时候slow才能指向删除节点的上一个节点（方便做删除操作），如图：
* fast和slow同时移动，直到fast指向末尾，如题：
* 删除slow指向的下一个节点，如图：
class Solution {
public :
ListNode * removeNthFromEnd ( ListNode * head , int n ) {
ListNode * dummyHead = new ListNode ( 0 );
dummyHead -> next = head ;
ListNode * slow = dummyHead ;
ListNode * fast = dummyHead ;
while ( n -- && fast != NULL ) {
fast = fast -> next ;
}
fast = fast -> next ; // fast再提前走一步，因为需要让slow指向删除节点的上一个节点
while ( fast != NULL ) {
fast = fast -> next ;
slow = slow -> next ;
}
slow -> next = slow -> next -> next ;
// ListNode *tmp = slow->next; C++释放内存的逻辑
// slow->next = tmp->next;
// delete nth;
return dummyHead -> next ;
}
};
public ListNode removeNthFromEnd ( ListNode head , int n ){
ListNode dummyNode = new ListNode ( 0 );
dummyNode . next = head ;
ListNode fastIndex = dummyNode ;
ListNode slowIndex = dummyNode ;
// 只要快慢指针相差 n 个结点即可
for ( int i = 0 ; i < n ; i ++ ){
fastIndex = fastIndex . next ;
}
while ( fastIndex != null ){
fastIndex = fastIndex . next ;
slowIndex = slowIndex . next ;
}
//此时 slowIndex 的位置就是待删除元素的前一个位置。
//具体情况可自己画一个链表长度为 3 的图来模拟代码来理解
slowIndex . next = slowIndex . next . next ;
return dummyNode . next ;
}
# 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 :
# 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
# 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
# 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
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func removeNthFromEnd ( head * ListNode , n int ) * ListNode {
dummyHead := & ListNode {}
dummyHead . Next = head
cur := head
prev := dummyHead
i := 1
for cur != nil {
cur = cur . Next
if i > n {
prev = prev . Next
}
i ++
}
prev . Next = prev . Next . Next
return dummyHead . Next
}
/**
* @param {ListNode} head
* @param {number} n
* @return {ListNode}
*/
var removeNthFromEnd = function ( head , n ) {
let ret = new ListNode ( 0 , head ),
slow = fast = ret ;
while ( n -- ) fast = fast . next ;
while ( fast . next !== null ) {
fast = fast . next ;
slow = slow . next
};
slow . next = slow . next . next ;
return ret . next ;
};
function removeNthFromEnd ( head : ListNode | null , n : number ) : ListNode | null {
let newHead : ListNode | null = new ListNode ( 0 , head );
//根据leetcode题目的定义可推断这里快慢指针均不需要定义为ListNode | null。
let slowNode : ListNode = newHead ;
let fastNode : ListNode = newHead ;
while ( n -- ) {
fastNode = fastNode . next ! ; //由虚拟头节点前进n个节点时,fastNode.next可推断不为null。
}
while ( fastNode . next ) { //遍历直至fastNode.next = null， 即尾部节点。 此时slowNode指向倒数第n个节点。
fastNode = fastNode . next ;
slowNode = slowNode . next ! ;
}
slowNode . next = slowNode . next ! . next ; //倒数第n个节点可推断其next节点不为空。
return newHead . next ;
}
function removeNthFromEnd ( head : ListNode | null , n : number ) : ListNode | null {
let curNode : ListNode | null = head ;
let listSize : number = 0 ;
while ( curNode ) {
curNode = curNode . next ;
listSize ++ ;
}
if ( listSize === n ) {
head = head . next ;
} else {
curNode = head ;
for ( let i = 0 ; i < listSize - n - 1 ; i ++ ) {
curNode = curNode . next ;
}
curNode . next = curNode . next . next ;
}
return head ;
};
function removeNthFromEnd ( head : ListNode | null , n : number ) : ListNode | null {
let newHead : ListNode | null = new ListNode ( 0 , head );
let cnt = 0 ;
function recur ( node ) {
if ( node === null ) return ;
recur ( node . next );
cnt ++ ;
if ( cnt === n + 1 ) {
node . next = node . next . next ;
}
}
recur ( newHead );
return newHead . next ;
};
fun removeNthFromEnd ( head : ListNode?, n : Int ): ListNode? {
val pre = ListNode ( 0 ). apply {
this . next = head
}
var fastNode : ListNode? = pre
var slowNode : ListNode? = pre
for ( i in 0. . n ) {
fastNode = fastNode ?. next
}
while ( fastNode != null ) {
slowNode = slowNode ?. next
fastNode = fastNode . next
}
slowNode ?. next = slowNode ?. next ?. next
return pre . next
}
func removeNthFromEnd ( _ head : ListNode ?, _ n : Int ) -> ListNode ? {
if head == nil {
return nil
}
if n == 0 {
return head
}
let dummyHead = ListNode ( - 1 , head )
var fast : ListNode ? = dummyHead
var slow : ListNode ? = dummyHead
// fast 前移 n
for _ in 0 .. < n {
fast = fast ?. next
}
while fast ?. next != nil {
fast = fast ?. next
slow = slow ?. next
}
slow ?. next = slow ?. next ?. next
return dummyHead . next
}
function removeNthFromEnd($head, $n) {
// 设置虚拟头节点
$dummyHead = new ListNode();
$dummyHead->next = $head;
$slow = $fast = $dummyHead;
while($n-- && $fast != null){
$fast = $fast->next;
}
// fast 再走一步，让 slow 指向删除节点的上一个节点
$fast = $fast->next;
while ($fast != NULL) {
$fast = $fast->next;
$slow = $slow->next;
}
$slow->next = $slow->next->next;
return $dummyHead->next;
}
object Solution {
def removeNthFromEnd ( head : ListNode , n : Int ): ListNode = {
val dummy = new ListNode ( - 1 , head ) // 定义虚拟头节点
var fast = head // 快指针从头开始走
var slow = dummy // 慢指针从虚拟头开始头
// 因为参数 n 是不可变量，所以不能使用 while(n>0){n-=1}的方式
for ( i <- 0 until n ) {
fast = fast . next
}
// 快指针和满指针一起走，直到fast走到null
while ( fast != null ) {
slow = slow . next
fast = fast . next
}
// 删除slow的下一个节点
slow . next = slow . next . next
// 返回虚拟头节点的下一个
dummy . next
}
}
impl Solution {
pub fn remove_nth_from_end ( head : Option < Box < ListNode >> , mut n : i32 ) -> Option < Box < ListNode >> {
let mut dummy_head = Box ::new ( ListNode ::new ( 0 ));
dummy_head . next = head ;
let mut fast = & dummy_head . clone ();
let mut slow = & mut dummy_head ;
while n > 0 {
fast = fast . next . as_ref (). unwrap ();
n -= 1 ;
}
while fast . next . is_some () {
fast = fast . next . as_ref (). unwrap ();
slow = slow . next . as_mut (). unwrap ();
}
slow . next = slow . next . as_mut (). unwrap (). next . take ();
dummy_head . next
}
}
/**c语言单链表的定义
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode * removeNthFromEnd ( struct ListNode * head , int n ) {
//定义虚拟头节点dummy 并初始化使其指向head
struct ListNode * dummy = malloc ( sizeof ( struct ListNode ));
dummy -> val = 0 ;
dummy -> next = head ;
//定义 fast slow 双指针
struct ListNode * fast = head ;
struct ListNode * slow = dummy ;
for ( int i = 0 ; i < n ; ++ i ) {
fast = fast -> next ;
}
while ( fast ) {
fast = fast -> next ;
slow = slow -> next ;
}
slow -> next = slow -> next -> next ; //删除倒数第n个节点
head = dummy -> next ;
free ( dummy ); //删除虚拟节点dummy
return head ;
}
public class Solution {
public ListNode RemoveNthFromEnd ( ListNode head , int n ) {
ListNode dummpHead = new ListNode ( 0 );
dummpHead . next = head ;
var fastNode = dummpHead ;
var slowNode = dummpHead ;
while ( n -- != 0 && fastNode != null )
{
fastNode = fastNode . next ;
}
while ( fastNode . next != null )
{
fastNode = fastNode . next ;
slowNode = slowNode . next ;
}
slowNode . next = slowNode . next . next ;
return dummpHead . next ;
}
}
此时不难写出如下C++代码：
