【题目】给定一个单链表的头节点head，请判断该链表是否为回文结构。
【例子】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。
【例子】如果链表长度为N，时间复杂度达到0（N），额外空间复杂度达到0（1）。

package Algorithms;

/**
 * @author : zhang
 * @version : 1.0
 * @date : Create in 2021/8/10
 * @description :
 */


import java.util.Stack;

public class IsPalindromeList {

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

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

    // need n extra space 笔试用此种方法
    public static boolean isPalindrome1(Node head) {
        Stack<Node> stack = new Stack<>();
        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;
    }

    // need n/2 extra space  快慢指针：cur走1步，right走两步
    public static boolean isPalindrome2(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;
    }

    // need O(1) extra space 面试用此方法回答（使用有限的几个变量解决）
    //过程举例：将 1 -> 2 -> 3 -> 2 -> 1 变为：1 -> 2 -> 3 <- 2 <- 1 进行比较
    //最后再变为1 -> 2 -> 3 -> 2 -> 1返回true
    public static boolean isPalindrome3(Node head) {
        if (head == null || head.next == null) {
            return true;
        }
        Node n1 = head;
        Node n2 = head;
        while (n2.next != null && n2.next.next != null) { // find mid node
            n1 = n1.next; // n1 -> mid
            n2 = n2.next.next; // n2 -> end
        }
        n2 = n1.next; // n2 -> right part first node
        n1.next = null; // mid.next -> null
        Node n3 = null;
        while (n2 != null) { // right part convert
            n3 = n2.next; // n3 -> save next node
            n2.next = n1; // next of right node convert
            n1 = n2; // n1 move
            n2 = n3; // n2 move
        }
        n3 = n1; // n3 -> save last node
        n2 = head;// n2 -> left first node
        boolean res = true;
        while (n1 != null && n2 != null) { // check palindrome
            if (n1.value != n2.value) {
                res = false;
                break;
            }
            n1 = n1.next; // left to mid
            n2 = n2.next; // right to mid
        }
        n1 = n3.next;
        n3.next = null;
        while (n1 != null) { // recover list
            n2 = n1.next;
            n1.next = n3;
            n3 = n1;
            n1 = n2;
        }
        return res;
    }

    public static void printLinkedList(Node node) {
        System.out.print("Linked List: ");
        while (node != null) {
            System.out.print(node.value + " ");
            node = node.next;
        }
        System.out.println();
    }

    public static void main(String[] args) {

        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.next.next.next = new Node(2);
        head.next.next.next.next = new Node(1);
        printLinkedList(head);
        System.out.print(isPalindrome1(head) + " | ");
        System.out.print(isPalindrome2(head) + " | ");
        System.out.println(isPalindrome3(head) + " | ");
        printLinkedList(head);

    }
}

/**
 * Linked List: 1 2 3 2 1 
 * true | true | true | 
 * Linked List: 1 2 3 2 1 
 */