反向后半链接列表
示例：
偶数：2->1->3->4->5->6->7->8====>2->1->3->4->8->7->6->5奇数：5->7->8->6->3->4->2====>5->7->8->2->4->3->6中间一个也需要反转
class ListNode
{
int val;
ListNode next;
ListNode(int x) { val = x; }
}
class ReverseRightHalfLinkedList
{
public static void main(String[] args)
{
ListNode node1 = new ListNode(1);
ListNode node2 = new ListNode(2);
ListNode node3 = new ListNode(3);
ListNode node4 = new ListNode(4);
ListNode node5 = new ListNode(5);
node1.next = node2;
node2.next = node3;
node3.next = node4;
node4.next = node5;
ListNode res = reverse(node1);//line 31
// ListNode node = node1;
// while (node != null)
// {
// System.out.println(node.val);
// node = node.next;
// }
}
public static ListNode reverse(ListNode start)
{
int counter = 0;
ListNode node = start;
ListNode pre = start;
while (node!= null)
{
counter += 1;
node = node.next;
}
for (int i=0; i<counter/2; i++)
{
pre = start;
start = start.next;
}
ListNode cur = start;
if (counter%2 ==0)
{
while (cur != null)
{
ListNode temp = cur.next;
cur.next = pre;
pre = cur;
cur = temp;
}
}
else
{
pre = pre.next;
cur = start.next;
System.out.println(pre.val);
System.out.println(cur.val);
while (cur != null)
{
ListNode temp = cur.next;
cur.next = pre;
pre = cur;
cur = temp;
System.out.println("-----");
System.out.println(pre.val); // line 90
System.out.println(cur.val);
System.out.println("-----");
System.out.println();
}
}
return start;
}
}
首先，我收到一条错误消息。
线程"main"java.lang.NullPointerException中的异常位于的ReverseRightHalfLinkedList.reverse(OA2.java:90(ReverseRightHalfLinkedList.main(OA2.java:31(
其次，我试着打印反向链表的顺序，它仍然是有序的。它没有被逆转。
请帮我解决这两个问题。非常感谢！
基于@激情的理念。我得到了一个更简洁的代码。
class ListNode
{
int val;
ListNode next;
ListNode(int x) { val = x; }
}
class ReverseRightHalfLinkedList
{
public static void main(String[] args)
{
ListNode node1 = new ListNode(2);
ListNode node2 = new ListNode(1);
ListNode node3 = new ListNode(3);
ListNode node4 = new ListNode(4);
ListNode node5 = new ListNode(5);
ListNode node6 = new ListNode(6);
ListNode node7 = new ListNode(7);
ListNode node8 = new ListNode(8);
node1.next = node2;
node2.next = node3;
node3.next = node4;
node4.next = node5;
node5.next = node6;
node6.next = node7;
node7.next = node8;
ListNode res = reverse(node1);
ListNode node = node1;
while (node != null)
{
System.out.println(node.val);
node = node.next;
}
}
public static ListNode reverse(ListNode start)
{
int counter = 0;
ListNode node = start;
ListNode pre = start;
ListNode result = start;
while (node!= null)// for count how many elements in linked list
{
counter += 1;
node = node.next;
}
for (int i=0; i< (counter / 2) ; i++)//no matter counter is even or odd, when it divided by 2, the result is even
{
pre = start;
start = start.next;
}
ListNode temp = null;
ListNode preNext = null;// this variable is used to track the next val behind pre
// for example, 2->1->3->4->5->6->7->8
// at this moment, pre:4, start:5
// I treated 5->6->7->8 as an independent linkedlist
// I reversed the linkedlist
// Finally, set the pre node's next value to the reversed linkedlist's head
// The first half and second half have been connected together
while (start != null)
{
temp = start.next;
start.next = preNext;
preNext = start;
start = temp;
}
pre.next = preNext;
return start;
}
}
您需要反转右半部分列表，因此您需要保存左半部分的最后一个节点和列表的标题，这样当右半部分被反转时，您可以将它们与左半部分链接并返回整个列表。
我已经改变了你的反向方法：
public static ListNode reverse(ListNode start)
{
int counter = 0;
ListNode node = start;
ListNode pre = start;
ListNode result = start;
while (node!= null)
{
counter += 1;
node = node.next;
}
int end = counter % 2 == 0 ? counter / 2 : (counter- 1) / 2 ;
for (int i=0; i< end ; i++)
{
pre = start;
start = start.next;
}
ListNode tlist = null,temp ;
while(start != null){
temp = start.next;
if(tlist == null){
tlist = start;
start.next = null;
}else{
start.next = tlist;
tlist = start;
}
start = temp;
}
pre.next = tlist;
return start;
}
希望这段代码能帮助您理解。
class MainClass {
public static void main(String args[]) {
Node head = new Node("Sameer");
//Adding data to my linked list
Node temp = addNode("Monica", head);
temp = addNode("Doug", temp);
temp = addNode("Eric", temp);
temp = addNode("Charlie", temp);
temp = addNode("Dan", temp);
temp = addNode("Enrique", temp);
temp = addNode("Ankitha", temp);
addNode("Chad", temp);
SolveMidLinkedList(head);
}
//method to add a node to the linked list
static Node addNode(String str, Node node) {
Node newNode = new Node(str);
node.link = newNode;
return newNode;
}
//method to reverse the right hald of the linkedlist
static void SolveMidLinkedList(Node head) {
int LinkedListSize = 0;
Node temp = head;
Node LastEle = null, MiddleEle = null, temp1 = null, temp2 = null;
//While loop to find the size of the linkedlist and also to find the last element in the linkedlist
while (temp != null) {
LastEle = temp;
temp = temp.link;
LinkedListSize++;
}
//Printing the names
temp = head;
System.out.println("Before rearranging the linked list");
while (temp != null) {
System.out.println(temp.name);
temp = temp.link;
}
//The main procedure
temp = head;
int iCount = 1;
while (temp != null) {
//Not changing the order of first half of the linked list
if (iCount <= (LinkedListSize / 2)) {
MiddleEle = temp;
} else {
//Reversing the order of second half(right side half) of the linked list.
temp2 = temp.link;
temp.link = temp1;
temp1 = temp;
}
temp = temp.link;
iCount++;
}
//At the end asssigning the middle element to the last element of the linked list.
MiddleEle.link = LastEle;
//Printing the names
temp = head;
System.out.println("After rearranging the linked list");
while (temp != null) {
System.out.println(temp.name);
temp = temp.link;
}
}
}
//General definition of Node in a linked list.
class Node {
Node link = null;
String name;
Node(String str) {
this.name = str;
}
}
/*
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* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
package javaapplication2;
class LinkedList {
static Node head;
static class Node {
int data;
Node next;
Node(int d) {
data = d;
next = null;
}
}
Node rev(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;
}
Node reverse(Node node) {
int count =0;
Node doo = node;
Node mid_pre = node;
Node mid = node;
while(doo.next != null){
if((count & 1) == 1){
// mid_pre = mid;
mid = mid.next;
}
doo = doo.next;
count++;
}
// System.out.println(" midd ddddd :"+mid_pre.data+" "+mid.data);
mid.next = rev(mid.next);
return node;
}
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(1);
list.head.next = new Node(2);
list.head.next.next = new Node(3);
list.head.next.next.next = new Node(4);
list.head.next.next.next.next = new Node(5);
list.head.next.next.next.next.next = new Node(6);
//// 1 2 3 4 5 6
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);
}
}
为什么不做一些非常简单的事情呢。我们知道名单的大小。我们可以使用list.get(counter)从末端到中心进行迭代。当列表大小是偶数或奇数但可行时，会有一些挑战。
public static void reverseFromEndToCentre() {
List list = new ArrayList<Integer>(Arrays.asList(1,2,3,4,5,6,7,8,9,10));
int midOfList = (int)Math.ceil(list.size() / 2);
int lenthOfList = list.size();
int counterFirstHalf =0;
while(counterFirstHalf<midOfList){
System.out.println(list.get(counterFirstHalf));
counterFirstHalf++;
}
int counterSecondHalf = lenthOfList;
while( counterSecondHalf > counterFirstHalf){
counterSecondHalf--;
System.out.println(list.get(counterSecondHalf ));
}
}
这里有一种反转链表后半部分(右(的方法。
#include<stdio.h>
#include<stdlib.h>
struct Node{
int data;
struct Node* next;
};
void append(struct Node** head_ref, int new_data)
{
struct Node* new_node=(struct Node*)malloc(sizeof(struct Node));
new_node->data=new_data;
new_node->next=NULL;
struct Node* last=*head_ref;
if(*head_ref==NULL){
*head_ref=new_node;
return;
}
while(last->next!=NULL)
last=last->next;
last->next=new_node;
return;
}
int display(struct Node* n)
{
int m=0;
printf("n");
while(n!=NULL)
{
printf(" %d ",n->data);
n=n->next;
m=m+1;
}
return m;
}
void reverse(struct Node** head_ref, int mid)
{
if(*head_ref==NULL)
{
printf("nEmpty List cannot be reversed");
return;
}
struct Node* last=*head_ref;
struct Node* second_last;
struct Node* beg=*head_ref;
int c=1;
while(c<=mid){
second_last=last;
last=last->next;
c=c+1;
}
struct Node* prev=NULL;
struct Node* current=last;
struct Node* next;
while(current != NULL)
{
next=current->next;
current->next=prev;
prev=current;
current=next;
}
*head_ref=beg;
second_last->next=prev;
}
int main()
{
int size;
struct Node* head=NULL;
int i,mid;
for(i=0;i<11;i++)
{
append(&head,rand()%19 +1);
}
size=display(head);
printf("n Size of linked list: %d",size);
if(size%2==0)
mid=(size+1)/2;
else
mid=size/2;
reverse(&head, mid);
display(head);
return 0;
}