[链表]

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;}