[链表] - 判断一个链表是否为回文结构

题目：
给定链表的头节点，判断该链表是否为会问结构
如果链表的长度为N，时间复杂度达到O(N),额外空间复杂度达到O(1)

方法一：

public class Node{
public int value;
public Node next;

public Node(int data){
this.value = data;
}
}

public boolean isPalindromel(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;
}

方法二：

public boolean isPalindromel2(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;
}
}
return true;
}

方法三：

public boolean isPalindromel3(Node head){
if(head == null || head.next == null){
return true;
}
Node n1 = head;
Node n2 = head;
while(n2.next != null && n2.next.next != null){
n1 = n1.next;
n2 = n2.next.next;
}
n2 = n1.next;
n1.next = null;
Node n3 = null;
while(n2 != null){
n3 = n2.next;
n2.next = n1;
n1 = n2;
n2 = n3;
}
n3 = n1;
n2 = head;
boolean res = true;
while(n1 != null && n2 != null){
if(n1.value != n2.value){
res = false;
break;
}
n1 = n1.next;
n2 = n2.next;
}
n1 = n3.next;
n3.next = null;
while(n1 != null){
n2 = n1.next;
n1.next = n3;
n3 = n1;
n1 = n2;
}
return res;
}