判断一个链表是否为回文结构

给定一个链表的头节点 head，请判断该链表是否为回文结构。

例如：

1->2->1，返回 true。

1->2->2->1，返回 true。

15->6->15，返回 true。

1->2->3，返回 false。
如果链表长度为 N，时间复杂度达到 O(N)，额外空间复杂度达到 O(1)

public class IsPalindromeList{
public static class Node{
public int value;
public Node next;
public Node (int data){
this.value = data;
}
}

//需要n额外空间
public static boolean isPalindromeList1(Node head){
Stack<Node> stack = new Stack<Node>();
Node cur = head;
while(cur != null){
stack.push(cur);
cur = cur.next;
}
while(head != null){
if(head.value != stack.pop().value){
return false;
}
head = head.next;
}
return true;
}

//需要n/2额外空间
public static boolean isPalindromeList2(Node head){
if(head == null || head.next == null){
return true;
}
Node right = head.next;
Node cur = head;
while(cur.next != null && cur.next.next != null){
right = right.next;
cur = cur.next.next;
}
Stack<Node> stack = new Stack<Node>();
while(right != null){
stack.push(right);
right = right.next;
}
while(!stack.isEmpty()){
if(head.value != stack.pop().value){
return false;
}
head = head.next;
}
return true;
}

//需要O(1)额外空间
public static boolean isPalindromeList(Node head){
if(head == null || head.next == null){
return true;
}
Node n1 = head;
Node n2 = head;
while(n2.next != null && n2.next.next != null){ //找到中间节点
n1 = n1.next; //n1指向中间节点
n2 = n2.next.next; //n2指向末尾节点
}
n2 = n1.next; //n2指向右部分第一个节点
n1.next = null; //中间节点下一节点指向空
Node n3 = null;
while(n2 != null){ //右部分链表翻转
n3 = n2.next; //n3用于保存下一节点
n2.next = n1; //右部分节点翻转
n1 = n2; //n1 移动
n2 = n3; //n2 移动
}
n3 = n1; //n3 用于保存最后节点
n2 = head; //n2指向左边第一个节点
boolean res = true;
while(n1 != null && n2 != null){
if(n1.value != n2.value){
res = false;
break;
}
n1 = n1.next; //左->中
n2 = n2.next; //右->中
}
n1 = n3.next;
n3.next = null;
while(n1 != null){
//恢复链表
n2 = n1.next;
n1.next = n3;
n3 = n1;
n1 = n2;
}
return res;
}
}