### 以递归方式反转Java中的链表

ListNode：

``````public class ListNode{
public String data;
public ListNode next;
}
``````

``````public AddressList reverse(){
}
``````

``````private AddressList reverse(ListNode current, AddressList reversedList){
if(current == null)
return reversedList;
return this.reverse(current.getNext(), reversedList);
}
``````

#### 32 个答案:

1. null（空列表）的反转是什么？空。
2. 单元素列表的反转是什么？元素。
3. n元素列表的反转是什么？与列表其余部分相反的是第一个元素。

4. ``````public ListNode Reverse(ListNode list)
{
if (list == null) return null; // first question

if (list.next == null) return list; // second question

// third question - in Lisp this is easy, but we don't have cons
// so we grab the second element (which will be the last after we reverse it)

ListNode secondElem = list.next;

// bug fix - need to unlink list from the rest or you will get a cycle
list.next = null;

// then we reverse everything from the second element on
ListNode reverseRest = Reverse(secondElem);

// then we join the two lists
secondElem.next = list;

return reverseRest;
}
``````

```1->2->3->4->5->null
it would become:
5->4->3->2->1->null```
``````
//Takes as parameters a node in a linked list, and p, the previous node in that list
//returns the head of the new list
Node reverse(Node n,Node p){
if(n==null) return null;
if(n.next==null){ //if this is the end of the list, then this is the new head
n.next=p;
return n;
}
Node r=reverse(n.next,n);  //call reverse for the next node,
//using yourself as the previous node
n.next=p;                     //Set your next node to be the previous node
return r;                     //Return the head of the new list
}
``````

``````Node reverse(Node head) {
// if head is null or only one node, it's reverse of itself.

// reverse the sub-list leaving the head node.

// head.next still points to the last element of reversed sub-list.
// so move the head to end.

// point last node to nil, (get rid of cycles)
return reverse;
}
``````

``````class Node<T>
{
Node<T> next;
public T data;
}

{

public void Reverse()
{
}

private Node<T> RecursiveReverse(Node<T> prev, Node<T> curr)
{
Node<T> next = curr.next;
curr.next = prev;
return (next == null) ? curr : RecursiveReverse(curr, next);
}
}
``````

• 跟踪头部
• 递归到链接列表的最后
• 反向连接

``````Head
|
1-->2-->3-->4-->N-->null

null-->1-->2-->3-->4-->N<--null

null-->1-->2-->3-->4<--N<--null

null-->1-->2-->3<--4<--N<--null

null-->1-->2<--3<--4<--N<--null

null-->1<--2<--3<--4<--N<--null

null<--1<--2<--3<--4<--N
|
``````

``````public ListNode reverse(ListNode toBeNextNode, ListNode currentNode)
{

if ((currentNode==null ||currentNode.next==null )&& toBeNextNode ==null)return currentHead; // ignore for size 0 & 1

if (currentNode.next!=null)currentHead = reverse(currentNode, currentNode.next); // travarse till end recursively

currentNode.next = toBeNextNode; // reverse link

}
``````

``````head-->12345

``````

``````// Example:
// reverse0(1->2->3, null) =>
//      reverse0(2->3, 1) =>
//          reverse0(3, 2->1) => reverse0(null, 3->2->1)
// once the first argument is null, return the second arg
// which is nothing but the reveresed list.

if (f != null) {
// now it has (n-1) elements
reverse0(f, t);
}
return n;
}
``````

``````public Node reverse(Node previous, Node current) {
if(previous == null)
return null;
previous.setNext(null);
if(current == null) {    // end of list
} else {
Node temp = current.getNext();
current.setNext(previous);
reverse(current, temp);
}
return null;    //should never reach here.
}
``````

``````Node newHead = reverse(head, head.getNext());
``````

``````public Node reverseListRecursive(Node curr)
{
if(curr == null){//Base case
}
else{
(reverseListRecursive(curr.next)).next = (curr);
}
return curr;
}
``````

```void reverse(node1,node2){
if(node1.next!=null)
reverse(node1.next,node1);
node1.next=node2;
}
call this method as reverse(start,null);
```

``````public void reverse() {
}

private Node reverseNodes(Node prevNode, Node currentNode) {
if (currentNode == null)
return prevNode;
Node nextNode = currentNode.next;
currentNode.next = prevNode;
return reverseNodes(currentNode, nextNode);
}
``````

``````public ListNode reverse(ListNode head) {
}
``````

``````public ListNode reverse(ListNode head) {
ListNode prev = null;
while (next != null) {
cur.next = prev;
prev = cur;
cur = next;
next = next.next;
}
return cur;
}
``````

``````public static ListNode recRev(ListNode curr){

if(curr.next == null){
return curr;
}
curr.next.next = curr;
curr.next = null;

}
``````

``````public static Node reverse(Node root) {
if (root == null || root.next == null) {
return root;
}

Node curr, prev, next;
curr = root; prev = next = null;
while (curr != null) {
next = curr.next;
curr.next = prev;

prev = curr;
curr = next;
}
return prev;
}
``````

``````public static Node reverseR(Node node) {
if (node == null || node.next == null) {
return node;
}

Node next = node.next;
node.next = null;

Node remaining = reverseR(next);
next.next = node;
return remaining;
}
``````

``````/**
* Reverse the list
* @return reference to the new list head
*/
if (next == null) {
return this; // Return the old tail of the list as the new head
}
LinkNode oldTail = next.reverse(); // Recurse to find the old tail
next.next = this; // The old next node now points back to this node
next = null; // Make sure old head has no next
return oldTail; // Return the old tail all the way back to the top
}
``````

``````public class LinkNode {
private char name;

/**
* Return a linked list of nodes, whose names are characters from the given string
* @param str node names
*/
if ((str == null) || (str.length() == 0)) {
throw new IllegalArgumentException("LinkNode constructor arg: " + str);
}
name = str.charAt(0);
if (str.length() > 1) {
}
}

public String toString() {
return name + ((next == null) ? "" : next.toString());
}

public static void main(String[] args) {
}
}
``````

PointZeroTwo得到了优雅的回答＆amp;在Java中也一样......

``````public void reverseList(){
}
}

private Node reverseListNodes(Node parent , Node child ){
Node next = child.next;
child.next = parent;
return (next==null)?child:reverseListNodes(child, next);
}
``````

``````static ListNode reverseR(ListNode head) {
}

// reverse the rest of the list recursively

// fix the first node after recursion
first.next.next = first;
first.next = null;

}
``````

an article启发，讨论递归数据结构的不可变实现，我使用Swift将替代解决方案放在一起。

1. nil（空列表）的反转是什么？
• 这里没关系，因为我们在Swift中没有保护。
2. 单元素列表的反转是什么？
• 元素本身
3. n元素列表的反转是什么？
• 第二个元素的反转，后跟第一个元素。
4. 我在下面的解决方案中适用了这些内容。

``````/**
Node is a class that stores an arbitrary value of generic type T
and a pointer to another Node of the same time.  This is a recursive
data structure representative of a member of a unidirectional linked
list.
*/
public class Node<T> {
public let value: T
public let next: Node<T>?

public init(value: T, next: Node<T>?) {
self.value = value
self.next = next
}

public func reversedList() -> Node<T> {
if let next = self.next {
// 3. The reverse of the second element on followed by the first element.
return next.reversedList() + value
} else {
// 2. Reverse of a one element list is itself
return self
}
}
}

/**
@return Returns a newly created Node consisting of the lhs list appended with rhs value.
*/
public func +<T>(lhs: Node<T>, rhs: T) -> Node<T> {
let tail: Node<T>?
if let next = lhs.next {
// The new tail is created recursively, as long as there is a next node.
tail = next + rhs
} else {
// If there is not a next node, create a new tail node to append
tail = Node<T>(value: rhs, next: nil)
}
// Return a newly created Node consisting of the lhs list appended with rhs value.
return Node<T>(value: lhs.value, next: tail)
}
``````

``````  public ListNode reverseR(ListNode p) {

//Base condition, Once you reach the last node,return p
if (p == null || p.next == null) {
return p;
}
//Go on making the recursive call till reach the last node,now head points to the last node

//Here, p points to the last but one node(previous node),  q points to the last   node. Then next next step is to adjust the links
ListNode q = p.next;

//Last node link points to the P (last but one node)
q.next = p;
//Set the last but node (previous node) next to null
p.next = null;
}
``````

``````public Node reverseRec(Node prev, Node curr) {
if (curr == null) return null;

if (curr.next == null) {
curr.next = prev;
return curr;

} else {
Node temp = curr.next;
curr.next = prev;
return reverseRec(curr, temp);
}
}
``````

``````public void reverseLinkedList(Node node){
if(node==null){
return;
}

Node temp = node.next;
node.next=node.prev;
node.prev=temp;
return;
}
``````

``````//Recursive solution
class SLL
{
int data;
SLL next;
}

{
//base case - 0 or 1 elements

return temp;
}
``````

``````package basic;

import custom.ds.nodes.Node;

private static Node<Integer> first = null;

public static void main(String[] args) {
Node<Integer> f = new Node<Integer>();
Node<Integer> s = new Node<Integer>();
Node<Integer> t = new Node<Integer>();
Node<Integer> fo = new Node<Integer>();
f.setNext(s);
s.setNext(t);
t.setNext(fo);
fo.setNext(null);

f.setItem(1);
s.setItem(2);
t.setItem(3);
fo.setItem(4);
Node<Integer> curr = f;
display(curr);
revLL(null, f);
display(first);
}

public static void display(Node<Integer> curr) {
while (curr.getNext() != null) {
System.out.println(curr.getItem());
System.out.println(curr.getNext());
curr = curr.getNext();
}
}

public static void revLL(Node<Integer> pn, Node<Integer> cn) {
while (cn.getNext() != null) {
revLL(cn, cn.getNext());
break;
}
if (cn.getNext() == null) {
first = cn;
}
cn.setNext(pn);
}
``````

}

``````public class Singlelinkedlist {
public static void main(String[] args) {
Elem list  = new Elem();
Reverse(list); //list is populate some  where or some how
}

//this  is the part you should be concerned with the function/Method has only 3 lines

public static void Reverse(Elem e){
if (e!=null)
if(e.next !=null )
Reverse(e.next);
//System.out.println(e.data);
}
}

class Elem {
public Elem next;    // Link to next element in the list.
public String data;  // Reference to the data.
}
``````

``````private Node ReverseList(Node current, Node previous)
{
if (current == null) return null;
Node originalNext = current.next;
current.next = previous;
if (originalNext == null) return current;
return ReverseList(originalNext, current);
}
``````

``````//this function reverses the linked list
public Node reverseList(Node p) {
return null;
}
//make the last node as head
if(p.next == null){
return p;
}
//traverse to the last node, then reverse the pointers by assigning the 2nd last node to last node and so on..
return reverseList(p.next).next = p;
}
``````

``````static void reverseList(){

ListNode Third=Second.next;
tail.next=null;//tail previous head is poiniting null
Second.next=tail;
ListNode current=Third;
ListNode prev=Second;
if(Third.next!=null){

while(current!=null){
ListNode    next=current.next;
current.next=prev;
prev=current;
current=next;
}
}
}
}
class ListNode{
public int data;
public ListNode next;
public int getData() {
return data;
}

public ListNode(int data) {
super();
this.data = data;
this.next=null;
}

public ListNode(int data, ListNode next) {
super();
this.data = data;
this.next = next;
}

public void setData(int data) {
this.data = data;
}
public ListNode getNext() {
return next;
}
public void setNext(ListNode next) {
this.next = next;
}

}
``````

``````Public LinkedList reverse(LinkedList List)
{
if(List == null)
return null;
if(List.next() == null)
return List;
LinkedList temp = this.reverse( List.next() );
return temp.setNext( List );
}
``````

``````package com.mypackage;
class list{

node first;
node last;

list(){
first=null;
last=null;
}

/*returns true if first is null*/
public boolean isEmpty(){
return first==null;
}
/*Method for insertion*/

public void insert(int value){

if(isEmpty()){
first=last=new node(value);
last.next=null;
}
else{
node temp=new node(value);
last.next=temp;
last=temp;
last.next=null;
}

}
/*simple traversal from beginning*/
public void traverse(){
node t=first;
while(!isEmpty() && t!=null){
t.printval();
t= t.next;
}
}
/*static method for creating a reversed linked list*/
public static void reverse(node n,list l1){

if(n.next!=null)
reverse(n.next,l1);/*will traverse to the very end*/
l1.insert(n.value);/*every stack frame will do insertion now*/

}
/*private inner class node*/
private class node{
int value;
node next;
node(int value){
this.value=value;
}
void printval(){
System.out.print(value+" ");
}
}

}
``````

``````List Reverse(List l)
{
if (IsEmpty(l) || Size(l) == 1) return l;
return reverse(rest(l))::first(l);
}
``````

rest（l）返回一个列表，该列表是没有第一个节点的原始列表。 first（l）返回第一个元素。 ::是一个连接运算符。

``````scala> import scala.collection.mutable.LinkedList

ll.foldLeft(LinkedList.empty[A])((accumulator, nextElement) => nextElement +: accumulator)

``````

``````public void reverse(){
if(isEmpty()){
return;
}
}

private Node<T> reverse(Node<T> node, Node<T> revHead){
if(node.next == null){
return node;
}
reverse.next = node;
node.next = null;
return node;
}
``````

``````    public void Reverse()
{
Node currentNode, nextNode=null, prevNode=null;