在 Java 中递归地反转链表

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I was asked this question at an interview and was annoyed that I fumbled with it since I was a little nervous.

This should reverse a singly linked list, called with reverse(head,NULL); so if this were your list:

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
}

edit: ive done like 6 edits on this, showing that it's still a little tricky for me lol

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最佳答案

There's code in one reply that spells it out, but you might find it easier to start from the bottom up, by asking and answering tiny questions (this is the approach in The Little Lisper):

What is the reverse of null (the empty list)? null.
What is the reverse of a one element list? the element.
What is the reverse of an n element list? the reverse of the rest of the list followed by the first element.

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;
}

小开

The algo will need to work on the following model,

keep track of the head
Recurse till end of linklist
Reverse linkage

Structure:

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
|
Head

Code:

public ListNode reverse(ListNode toBeNextNode, ListNode currentNode)
{
ListNode currentHead = currentNode; // keep track of the head


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


return currentHead;
}

Output:

head-->12345


head-->54321

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I got half way through (till null, and one node as suggested by plinth), but lost track after making recursive call. However, after reading the post by plinth, here is what I came up with:

Node reverse(Node head) {
// if head is null or only one node, it's reverse of itself.
if ( (head==null) || (head.next == null) ) return head;


// reverse the sub-list leaving the head node.
Node reverse = reverse(head.next);


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


// point last node to nil, (get rid of cycles)
head.next = null;
return reverse;
}

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

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I think this is more cleaner solution, which resembles LISP

// 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.


Link reverse0(Link f, Link n) {
if (f != null) {
Link t = new Link(f.data1, f.data2);
t.nextLink = n;
f = f.nextLink;             // assuming first had n elements before,
// now it has (n-1) elements
reverse0(f, t);
}
return n;
}

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I know this is an old post, but most of the answers are not tail recursive i.e. they do some operations after returning from the recursive call, and hence not the most efficient.

Here is a tail recursive version:

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

Call with:

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

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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.
}

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

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Here's yet another recursive solution. It has less code within the recursive function than some of the others, so it might be a little faster. This is C# but I believe Java would be very similar.

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


class LinkedList<T>
{
Node<T> head = null;


public void Reverse()
{
if (head != null)
head = RecursiveReverse(null, head);
}


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);
}
}

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public static ListNode recRev(ListNode curr){


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


// propogate the head value
return head;


}

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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);
}
}

call using: head = reverseRec(null, head);

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PointZeroTwo has got elegant answer & the same in Java ...

public void reverseList(){
if(head!=null){
head = reverseListNodes(null , head);
}
}


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

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What other guys done , in other post is a game of content, what i did is a game of linkedlist, it reverse the LinkedList's member not reverse of a Value of members.

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 );
}

小开

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+" ");
}
}


}

小开

Reverse by recursive algo.

public ListNode reverse(ListNode head) {
if (head == null || head.next == null) return head;
ListNode rHead = reverse(head.next);
rHead.next = head;
head = null;
return rHead;
}

By iterative

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

小开

public void reverse(){
if(isEmpty()){
return;
}
Node<T> revHead = new Node<T>();
this.reverse(head.next, revHead);
this.head = revHead;
}


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

小开

public void reverse() {
head = reverseNodes(null, head);
}


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

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Here is a reference if someone is looking for Scala implementation:

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


scala> def reverseLinkedList[A](ll: LinkedList[A]): LinkedList[A] =
ll.foldLeft(LinkedList.empty[A])((accumulator, nextElement) => nextElement +: accumulator)
reverseLinkedList: [A](ll: scala.collection.mutable.LinkedList[A])scala.collection.mutable.LinkedList[A]


scala> reverseLinkedList(LinkedList("a", "b", "c"))
res0: scala.collection.mutable.LinkedList[java.lang.String] = LinkedList(c, b, a)


scala> reverseLinkedList(LinkedList("1", "2", "3"))
res1: scala.collection.mutable.LinkedList[java.lang.String] = LinkedList(3, 2, 1)

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This solution demonstrates that no arguments are required.

/**
* Reverse the list
* @return reference to the new list head
*/
public LinkNode reverse() {
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
}

Here is the supporting code, to demonstrate that this works:

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


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


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


public static void main(String[] args) {
LinkNode head = new LinkNode("abc");
System.out.println(head);
System.out.println(head.reverse());
}
}

小开

Here is a simple iterative approach:

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;
}

And here is a recursive approach:

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;
}

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The solution is:

package basic;


import custom.ds.nodes.Node;


public class RevLinkedList {


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);
}

}

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static void reverseList(){


if(head!=null||head.next!=null){
ListNode tail=head;//head points to tail




ListNode Second=head.next;
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;
}
}
head=prev;//new head
}
}
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;
}










}

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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);
}

小开

As Java is always pass-by-value, to recursively reverse a linked list in Java, make sure to return the "new head"(the head node after reversion) at the end of the recursion.

static ListNode reverseR(ListNode head) {
if (head == null || head.next == null) {
return head;
}


ListNode first = head;
ListNode rest = head.next;


// reverse the rest of the list recursively
head = reverseR(rest);


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


return head;
}

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//this function reverses the linked list
public Node reverseList(Node p) {
if(head == null){
return null;
}
//make the last node as head
if(p.next == null){
head.next = null;
head = p;
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;
}

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This is how we would do this in Opal - a pure functional programming language. And, IMHO - doing this recursively only makes sense in that context.

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

rest(l) returns a list that is the original list without it's first node. first(l) returns the first element. :: is a concatenation operator.

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


SLL reverse(SLL head)
{
//base case - 0 or 1 elements
if(head == null || head.next == null) return head;


SLL temp = reverse(head.next);
head.next.next = head;
head.next = null;
return temp;
}

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Inspired by an article discussing immutable implementations of recursive data structures I put an alternate solution together using Swift.

The leading answer documents solution by highlighting the following topics:

What is the reverse of nil (the empty list)?
- Does not matter here, because we have nil protection in Swift.
What is the reverse of a one element list?
- The element itself
What is the reverse of an n element list?
- The reverse of the second element on followed by the first element.

I have called these out where applicable in the solution below.

/**
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)
}

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Reversing the linked list using recursion. The idea is adjusting the links by reversing the links.

  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


ListNode head  = reverseR(p.next);  //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;
return head; //Head points to the last node
}

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Here is C# version of Reverse for linklist.

    public void Reverse()
{
Node currentNode, nextNode=null, prevNode=null;
currentNode = head;
while(currentNode!=null)
{
nextNode = currentNode.next;
currentNode.next = prevNode;
prevNode = currentNode;
currentNode = nextNode;
}
head = prevNode;
}

小开

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


reverseLinkedList(node.next);
Node temp = node.next;
node.next=node.prev;
node.prev=temp;
return;
}

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Solution in javascript (recursive) :

function reverse_linked_list_1(node){
function reverse_linked_list_1(node, result){
return node ? reverse_linked_list_1(node.next, {data: node.data, next: result}) : result;
}
return reverse_linked_list_1(node, null);
}