[数据结构与算法]链表面试题之判断是否为回文链表—三种方式解决，总有一种方法让面试官满意

	座右铭：代码虐我千百遍，我待代码如初恋！！！

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。

• 栈方法特别简单（笔试用)
• 改原链表的方法就需要注意边界了（面试用）

第一种方法：用一个栈来辅助实现，将链表从头结点开始依次压入栈中，之后再从栈顶依次弹出并与原链表一个一个进行比较，，如果相比较的过程全部相同，说明是回文结构，如果有一个比对不上就不是回文结构。因为栈的特征是先进后出的，所以弹出的过程就是逆序的过程。这个方法比较简单，缺点就是费点空间，需要O(N)的空间复杂度

第二中方法：第一种基础上改进一下，同样需要一个栈来辅助实现，但是只需要用到栈的一半空间，所以空间复杂度为O(N/2)。额外申请两个变量，一个快指针和慢指针。让慢指针指向链表右半部分的第一个结点，然后将右半部分压入栈中，再从栈中弹出和左半部分一一比较

第三中方法：不用申请栈，只需要O(1)的空间复杂度。申请一个快指针和慢指针，链表结点个数为奇数的时候慢指针来到中点位置，；偶数个的时候，慢指针来到上中点位置。然后将右半部分链表反转，慢指针指向的结点指向空。此时再让慢指针和快指针分别从最右边和最左边开始一一进行比较。最后返回结果之前一定要将右半部分链表反转回来！！！此方法难点在于难抠边界，但是是最优解，适合面试的时候和面试官聊聊哈哈哈~~~///(^v)\\\~~~

package com.harrison.class06;

import java.util.Stack;

public class Code02_IsPalindromeList {
	public static class Node {
		public int value;
		public Node next;

		public Node(int v) {
			value = v;
		}
	}

	// need n extra space
	public static boolean isPalindrome1(Node head) {
		Stack<Node> stack = new Stack<Node>();
		Node cur = head;
		// 将链表从头结点开始依次压入栈中
		while (cur != null) {
			stack.push(cur);
			cur = cur.next;
		}
		// 从栈顶依次弹出和链表开始比较
		while (head != null) {
			if (head.value != stack.pop().value) {
				return false;
			}
			head = head.next;
		}
		return true;
	}

	// need n/2 extra space
	public static boolean isPalindrome2(Node head) {
		if (head == null || head.next == null) {
			return true;
		}
		Node right = head.next;// 慢指针
		Node cur = head;// 快指针
		while (cur.next != null && cur.next.next != null) {
			right = right.next;
			cur = cur.next.next;
		}
		// right 奇数个来到中点位置 偶数个来到下中点位置
		Stack<Node> stack = new Stack<>();
		// 链表的右半部分依次压入栈中
		while (right != null) {
			stack.push(right);
			right = right.next;
		}
		// 从栈中依次弹出 和链表左部分依次比较
		while (!stack.isEmpty()) {
			if (head.value != stack.pop().value) {
				return false;
			}
			head = head.next;
		}
		return true;
	}

	// need O(1) extra space
	public static boolean isPalindrome3(Node head) {
		if (head == null || head.next == null) {
			return true;
		}
		Node n1 = head;// 慢指针 n1 -> mid
		Node n2 = head;// 快指针 n2 -> end
		while (n2.next != null && n2.next.next != null) {
			n1 = n1.next;
			n2 = n2.next.next;
		}
		// 慢指针来到中点或者上中点位置 快指针来到终点
		n2 = n1.next;// n2 来到右部分第一个结点
		n1.next = null;// n1.next -> null
		Node n3 = null;
		// 右部分反转
		while (n2 != null) {
			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;
		// 慢指针从右边开始 快指针从头结点开始 一一比较，任何一个为空就停下来
		while (n1 != null && n2 != null) {
			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;
		// 将右部分逆序回来
		while (n1 != null) {
			n2 = n1.next;
			n1.next = n3;
			n3 = n1;
			n1 = n2;
		}
		return res;
	}

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

	public static void main(String[] args) {

		Node head = null;
		printLinkedList(head);
		System.out.print(isPalindrome1(head) + " | ");
		System.out.print(isPalindrome2(head) + " | ");
		System.out.println(isPalindrome3(head) + " | ");
		printLinkedList(head);
		System.out.println("=========================");

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

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

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

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

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

		head = new Node(1);
		head.next = new Node(2);
		head.next.next = new Node(3);
		head.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);
		System.out.println("=========================");

		head = new Node(1);
		head.next = new Node(2);
		head.next.next = new Node(2);
		head.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);
		System.out.println("=========================");

		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);
		System.out.println("=========================");
	}
}