问题
Can someone tell me why my code dosent work? I want to reverse a single linked list in java: This is the method (that doesnt work correctly)
public void reverseList(){
Node before = null;
Node tmp = head;
Node next = tmp.next;
while(tmp != null){
if(next == null)
return;
tmp.next = before;
before = tmp;
tmp = next;
next = next.next;
}
}
And this is the Node class:
public class Node{
public int data;
public Node next;
public Node(int data, Node next){
this.data = data;
this.next = next;
}
}
On input 4->3->2->1 I got output 4. I debugged it and it sets pointers correctly but still I dont get why it outputs only 4.
回答1:
Node next = tmp.next;
while(tmp != null){
So what happens when tmp == null?
You almost got it, though.
Node before = null;
Node tmp = head;
while (tmp != null) {
Node next = tmp.next;
tmp.next = before;
before = tmp;
tmp = next;
}
head = before;
Or in nicer (?) naming:
Node reversedPart = null;
Node current = head;
while (current != null) {
Node next = current.next;
current.next = reversedPart;
reversedPart = current;
current = next;
}
head = reversedPart;
ASCII art:
<__<__<__ __ : reversedPart : head
(__)__ __ __
head : current: > > >
回答2:
public Node<E> reverseList(Node<E> node) {
if (node == null || node.next == null) {
return node;
}
Node<E> currentNode = node;
Node<E> previousNode = null;
Node<E> nextNode = null;
while (currentNode != null) {
nextNode = currentNode.next;
currentNode.next = previousNode;
previousNode = currentNode;
currentNode = nextNode;
}
return previousNode;
}
回答3:
The method for reversing a linked list is as below;
Reverse Method
public void reverseList() {
Node<E> curr = head;
Node<E> pre = null;
Node<E> incoming = null;
while(curr != null) {
incoming = curr.next; // store incoming item
curr.next = pre; // swap nodes
pre = curr; // increment also pre
curr = incoming; // increment current
}
head = pre; // pre is the latest item where
// curr is null
}
Three references are needed to reverse a list: pre, curr, incoming
... pre curr incoming
... --> (n-1) --> (n) --> (n+1) --> ...
To reverse a node, you have to store previous element, so that you can use the simple stament;
curr.next = pre;
To reverse the current element's direction. However, to iterate over the list, you have to store incoming element before the execution of the statement above because as reversing the current element's next reference, you don't know the incoming element anymore, that's why a third reference needed.
The demo code is as below;
LinkedList Sample Class
public class LinkedList<E> {
protected Node<E> head;
public LinkedList() {
head = null;
}
public LinkedList(E[] list) {
this();
addAll(list);
}
public void addAll(E[] list) {
for(int i = 0; i < list.length; i++)
add(list[i]);
}
public void add(E e) {
if(head == null)
head = new Node<E>(e);
else {
Node<E> temp = head;
while(temp.next != null)
temp = temp.next;
temp.next = new Node<E>(e);
}
}
public void reverseList() {
Node<E> curr = head;
Node<E> pre = null;
Node<E> incoming = null;
while(curr != null) {
incoming = curr.next; // store incoming item
curr.next = pre; // swap nodes
pre = curr; // increment also pre
curr = incoming; // increment current
}
head = pre; // pre is the latest item where
// curr is null
}
public void printList() {
Node<E> temp = head;
System.out.print("List: ");
while(temp != null) {
System.out.print(temp + " ");
temp = temp.next;
}
System.out.println();
}
public static class Node<E> {
protected E e;
protected Node<E> next;
public Node(E e) {
this.e = e;
this.next = null;
}
@Override
public String toString() {
return e.toString();
}
}
}
Test Code
public class ReverseLinkedList {
public static void main(String[] args) {
Integer[] list = { 4, 3, 2, 1 };
LinkedList<Integer> linkedList = new LinkedList<Integer>(list);
linkedList.printList();
linkedList.reverseList();
linkedList.printList();
}
}
Output
List: 4 3 2 1
List: 1 2 3 4
回答4:
If this isn't homework and you are doing this "manually" on purpose, then I would recommend using
Collections.reverse(list);
Collections.reverse() returns void, and your list is reversed after the call.
回答5:
We can have three nodes previous,current and next.
public void reverseLinkedlist()
{
/*
* Have three nodes i.e previousNode,currentNode and nextNode
When currentNode is starting node, then previousNode will be null
Assign currentNode.next to previousNode to reverse the link.
In each iteration move currentNode and previousNode by 1 node.
*/
Node previousNode = null;
Node currentNode = head;
while (currentNode != null)
{
Node nextNode = currentNode.next;
currentNode.next = previousNode;
previousNode = currentNode;
currentNode = nextNode;
}
head = previousNode;
}
回答6:
public void reverse() {
Node prev = null; Node current = head; Node next = current.next;
while(current.next != null) {
current.next = prev;
prev = current;
current = next;
next = current.next;
}
current.next = prev;
head = current;
}
回答7:
I know the recursive solution is not the optimal one, but just wanted to add one here:
public class LinkedListDemo {
static class Node {
int val;
Node next;
public Node(int val, Node next) {
this.val = val;
this.next = next;
}
@Override
public String toString() {
return "" + val;
}
}
public static void main(String[] args) {
Node n = new Node(1, new Node(2, new Node(3, new Node(20, null))));
display(n);
n = reverse(n);
display(n);
}
static Node reverse(Node n) {
Node tail = n;
while (tail.next != null) {
tail = tail.next;
}
reverseHelper(n);
return (tail);
}
static Node reverseHelper(Node n) {
if (n.next != null) {
Node reverse = reverseHelper(n.next);
reverse.next = n;
n.next = null;
return (n);
}
return (n);
}
static void display(Node n) {
for (; n != null; n = n.next) {
System.out.println(n);
}
}
}
回答8:
I don't get it... why not doing this :
private LinkedList reverseLinkedList(LinkedList originalList){
LinkedList reversedList = new LinkedList<>();
for(int i=0 ; i<originalList.size() ; i++){
reversedList.add(0, originalList.get(i));
}
return reversedList;
}
I find this easier.
回答9:
A more elegant solution would be to use recursion
void ReverseList(ListNode current, ListNode previous) {
if(current.Next != null)
{
ReverseList(current.Next, current);
ListNode temp = current.Next;
temp.Next = current;
current.Next = previous;
}
}
回答10:
I tried the below code and it works fine:
Node head = firstNode;
Node current = head;
while(current != null && current.next != null){
Node temp = current.next;
current.next = temp.next;
temp.next = head;
head = temp;
}
Basically one by one it sets the next pointer of one node to its next to next node, so from next onwards all nodes are attached at the back of the list.
回答11:
// Java program for reversing the linked list
class LinkedList {
static Node head;
static class Node {
int data;
Node next;
Node(int d) {
data = d;
next = null;
}
}
// Function to reverse the linked list
Node reverse(Node node) {
Node prev = null;
Node current = node;
Node next = null;
while (current != null) {
next = current.next;
current.next = prev;
prev = current;
current = next;
}
node = prev;
return node;
}
// prints content of double linked list
void printList(Node node) {
while (node != null) {
System.out.print(node.data + " ");
node = node.next;
}
}
public static void main(String[] args) {
LinkedList list = new LinkedList();
list.head = new Node(85);
list.head.next = new Node(15);
list.head.next.next = new Node(4);
list.head.next.next.next = new Node(20);
System.out.println("Given Linked list");
list.printList(head);
head = list.reverse(head);
System.out.println("");
System.out.println("Reversed linked list ");
list.printList(head);
}
}
OUTPUT: -
Given Linked list
85 15 4 20
Reversed linked list
20 4 15 85
回答12:
Node reverse_rec(Node start) {
if (start == null || start -> next == null) {
return start;
}
Node new_start = reverse(start->next);
start->next->next = start;
start->next = null;
return new_start;
}
Node reverse(Node start) {
Node cur = start;
Node bef = null;
while (cur != null) {
Node nex = cur.next;
cur.next = bef;
bef = cur;
cur = nex;
}
return bef;
}
回答13:
I think your problem is that your initially last element next attribute isn't being changed becuase of your condition
if(next == null)
return;
Is at the beginning of your loop.
I would move it right after tmp.next has been assigned:
while(tmp != null){
tmp.next = before;
if(next == null)
return;
before = tmp;
tmp = next;
next = next.next;
}
回答14:
Use this.
if (current== null || current.next==null) return current;
Node nextItem = current.next;
current.next = null;
Node reverseRest = reverse(nextItem);
nextItem.next = current;
return reverseRest
or Java Program to reverse a Singly Linked List
回答15:
package com.three;
public class Link {
int a;
Link Next;
public Link(int i){
a=i;
}
}
public class LinkList {
Link First = null;
public void insertFirst(int a){
Link objLink = new Link(a);
objLink.Next=First;
First = objLink;
}
public void displayLink(){
Link current = First;
while(current!=null){
System.out.println(current.a);
current = current.Next;
}
}
public void ReverseLink(){
Link current = First;
Link Previous = null;
Link temp = null;
while(current!=null){
if(current==First)
temp = current.Next;
else
temp=current.Next;
if(temp==null){
First = current;
//return;
}
current.Next=Previous;
Previous=current;
//System.out.println(Previous);
current = temp;
}
}
public static void main(String args[]){
LinkList objLinkList = new LinkList();
objLinkList.insertFirst(1);
objLinkList.insertFirst(2);
objLinkList.insertFirst(3);
objLinkList.insertFirst(4);
objLinkList.insertFirst(5);
objLinkList.insertFirst(6);
objLinkList.insertFirst(7);
objLinkList.insertFirst(8);
objLinkList.displayLink();
System.out.println("-----------------------------");
objLinkList.ReverseLink();
objLinkList.displayLink();
}
}
回答16:
You can also try this
LinkedListNode pointer = head;
LinkedListNode prev = null, curr = null;
/* Pointer variable loops through the LL */
while(pointer != null)
{
/* Proceed the pointer variable. Before that, store the current pointer. */
curr = pointer; //
pointer = pointer.next;
/* Reverse the link */
curr.next = prev;
/* Current becomes previous for the next iteration */
prev = curr;
}
System.out.println(prev.printForward());
回答17:
package LinkedList;
import java.util.LinkedList;
public class LinkedListNode {
private int value;
private LinkedListNode next = null;
public LinkedListNode(int i) {
this.value = i;
}
public LinkedListNode addNode(int i) {
this.next = new LinkedListNode(i);
return next;
}
public LinkedListNode getNext() {
return next;
}
@Override
public String toString() {
String restElement = value+"->";
LinkedListNode newNext = getNext();
while(newNext != null)
{restElement = restElement + newNext.value + "->";
newNext = newNext.getNext();}
restElement = restElement +newNext;
return restElement;
}
public static void main(String[] args) {
LinkedListNode headnode = new LinkedListNode(1);
headnode.addNode(2).addNode(3).addNode(4).addNode(5).addNode(6);
System.out.println(headnode);
headnode = reverse(null,headnode,headnode.getNext());
System.out.println(headnode);
}
private static LinkedListNode reverse(LinkedListNode prev, LinkedListNode current, LinkedListNode next) {
current.setNext(prev);
if(next == null)
return current;
return reverse(current,next,next.getNext());
}
private void setNext(LinkedListNode prev) {
this.next = prev;
}
}
回答18:
public class ReverseLinkedList {
public static void main(String args[]){
LinkedList<String> linkedList = new LinkedList<String>();
linkedList.add("a");
linkedList.add("b");
linkedList.add("c");
linkedList.add("d");
linkedList.add("e");
linkedList.add("f");
System.out.println("Original linkedList:");
for(int i = 0; i <=linkedList.size()-1; i++){
System.out.println(" - "+ linkedList.get(i));
}
LinkedList<String> reversedlinkedList = reverse(linkedList);
System.out.println("Reversed linkedList:");
for(int i = 0; i <=reversedlinkedList.size()-1; i++){
System.out.println(" - "+ reversedlinkedList.get(i));
}
}
public static LinkedList<String> reverse(LinkedList<String> linkedList){
for(int i = 0; i < linkedList.size()/2; i++){
String temp = linkedList.get(i);
linkedList.set(i, linkedList.get(linkedList.size()-1-i));
linkedList.set((linkedList.size()-1-i), temp);
}
return linkedList;
}
}
回答19:
To reverse a singly linked list you should have three nodes, top, beforeTop and AfterTop. Top is the header of singly linked list, hence beforeTop would be null and afterTop would be next element of top and with each iteration move forward beforeTop is assigned top and top is assigned afterTop(i.e. top.next).
private static Node inverse(Node top) {
Node beforeTop=null, afterTop;
while(top!=null){
afterTop=top.next;
top.next=beforeTop;
beforeTop=top;
top=afterTop;
}
return beforeTop;
}
回答20:
Using Recursion It's too easy :
package com.config;
import java.util.Scanner;
public class Help {
public static void main(String args[]){
Scanner sc = new Scanner(System.in);
Node head = null;
Node temp = null;
int choice = 0;
boolean flage = true;
do{
Node node = new Node();
System.out.println("Enter Node");
node.data = sc.nextInt();
if(flage){
head = node;
flage = false;
}
if(temp!=null)
temp.next = node;
temp = node;
System.out.println("Enter 0 to exit.");
choice = sc.nextInt();
}while(choice!=0);
Help.getAll(head);
Node reverse = Help.reverse(head,null);
//reverse = Help.reverse(head, null);
Help.getAll(reverse);
}
public static void getAll(Node head){
if(head==null)
return ;
System.out.println(head.data+"Memory Add "+head.hashCode());
getAll(head.next);
}
public static Node reverse(Node head,Node tail){
Node next = head.next;
head.next = tail;
return (next!=null? reverse(next,head) : head);
}
}
class Node{
int data = 0;
Node next = null;
}
回答21:
Node Reverse(Node head) {
Node n,rev;
rev = new Node();
rev.data = head.data;
rev.next = null;
while(head.next != null){
n = new Node();
head = head.next;
n.data = head.data;
n.next = rev;
rev = n;
n=null;
}
return rev;
}
Use above function to reverse single linked list.
回答22:
public ListNode reverseList(ListNode head) {
ListNode prev = null;
ListNode curr = head;
while (curr != null) {
ListNode nextTemp = curr.next;
curr.next = prev;
prev = curr;
curr = nextTemp;
}
return prev;
}
check more details about complexity analysis http://javamicro.com/ref-card/DS-Algo/How-to-Reverse-Singly-Linked-List?
回答23:
public static LinkedList reverseLinkedList(LinkedList node) {
if (node == null || node.getNext() == null) {
return node;
}
LinkedList remaining = reverseLinkedList(node.getNext());
node.getNext().setNext(node);
node.setNext(null);
return remaining;
}
回答24:
/**
* Reverse LinkedList
* @author asharda
*
*/
class Node
{
int data;
Node next;
Node(int data)
{
this.data=data;
}
}
public class ReverseLinkedList {
static Node root;
Node temp=null;
public void insert(int data)
{
if(root==null)
{
root=new Node(data);
}
else
{
temp=root;
while(temp.next!=null)
{
temp=temp.next;
}
Node newNode=new Node(data);
temp.next=newNode;
}
}//end of insert
public void display(Node head)
{
while(head!=null)
{
System.out.println(head.data);
head=head.next;
}
}
public Node reverseLinkedList(Node head)
{
Node newNode;
Node tempr=null;
while(head!=null)
{
newNode=new Node(head.data);
newNode.next=tempr;
tempr=newNode;
head=head.next;
}
return tempr;
}
public static void main(String[] args) {
ReverseLinkedList r=new ReverseLinkedList();
r.insert(10);
r.insert(20);
r.insert(30);
r.display(root);
Node t=r.reverseLinkedList(root);
r.display(t);
}
}
回答25:
public class SinglyLinkedListImpl<T> {
private Node<T> head;
public void add(T element) {
Node<T> item = new Node<T>(element);
if (head == null) {
head = item;
} else {
Node<T> temp = head;
while (temp.next != null) {
temp = temp.next;
}
temp.next = item;
}
}
private void reverse() {
Node<T> temp = null;
Node<T> next = null;
while (head != null) {
next = head.next;
head.next = temp;
temp = head;
head = next;
}
head = temp;
}
void printList(Node<T> node) {
while (node != null) {
System.out.print(node.data + " ");
node = node.next;
}
System.out.println();
}
public static void main(String a[]) {
SinglyLinkedListImpl<Integer> sl = new SinglyLinkedListImpl<Integer>();
sl.add(1);
sl.add(2);
sl.add(3);
sl.add(4);
sl.printList(sl.head);
sl.reverse();
sl.printList(sl.head);
}
static class Node<T> {
private T data;
private Node<T> next;
public Node(T data) {
super();
this.data = data;
}
}
}
回答26:
public class Linkedtest {
public static void reverse(List<Object> list) {
int lenght = list.size();
for (int i = 0; i < lenght / 2; i++) {
Object as = list.get(i);
list.set(i, list.get(lenght - 1 - i));
list.set(lenght - 1 - i, as);
}
}
public static void main(String[] args) {
LinkedList<Object> st = new LinkedList<Object>();
st.add(1);
st.add(2);
st.add(3);
st.add(4);
st.add(5);
Linkedtest.reverse(st);
System.out.println("Reverse Value will be:"+st);
}
}
This will be useful for any type of collection Object.
来源：https://stackoverflow.com/questions/22605050/reverse-singly-linked-list-java