判断一个链表是否为回文结构【题目】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。进阶：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)。

方式1：借助栈空间辅助度是O(N)

方式2：借助栈空间复杂度是 O(n/2)。只存后半个链表

方式3：反转后半个链表最后再反转回来

  1 package my_basic.class_3;
  2 
  3 import java.util.Stack;
  4 
  5 //是否是回文结构  121  1221，
  6 public class Code_11_IsPalindromeList {
  7     public static class Node{
  8         int value;
  9         Node next;
 10         public Node(int value) {
 11             super();
 12             this.value = value;
 13         }
 14     }
 15     
 16     //need extra(n) space 栈辅助
 17     public static boolean isPalindrome1(Node head) {
 18         Stack<Node> stack = new Stack<Node>();
 19         Node cur = head;
 20         while(cur != null) {
 21             stack.push(cur);
 22             cur = cur.next;
 23         }
 24         while (head != null && !stack.empty()) {
 25             if (head.value != stack.pop().value) {
 26                 return false;
 27             }
 28             head = head.next;
 29         }
 30         if (head==null) {
 31             System.out.println("head null");
 32         }
 33         return true;
 34     }
 35     
 36     //need n/2 extra space 栈辅助
 37     public static Boolean isPalindrome2(Node head) {
 38         Stack<Node> stack = new Stack<Node>();
 39         if (head == null || head.next == null) {
 40             return true;
 41         }
 42         Node right = head.next;
 43         Node cur = head;
 44         if (cur.next!=null || cur.next.next !=null) {
 45             cur = cur.next.next;
 46             right = right.next;
 47         }
 48         while (right != null) {
 49             stack.push(right);
 50             right = right.next;
 51         }
 52         while(!stack.empty()) {
 53             if (head.value != stack.pop().value) {
 54                 return false;
 55             }
 56             head = head.next;
 57         }
 58         return true;
 59     }
 60     
 61     //need O(1) extra space 反转半个链表
 62     public static Boolean isPalindrome3(Node head) {
 63         if (head == null || head.next == null) {
 64             return true;
 65         }
 66         Node n1 = head;
 67         Node n2 = head;
 68         while (n2.next != null && n2.next.next!=null) {
 69             n1 = n1.next;  //mid
 70             n2 = n2.next.next; //last
 71         }
 72         n2 = n1.next;  //右边第一个节点
 73         n1.next = null;
 74         Node n3 = null;
 75         while (n2 != null) {   /*反转右半部分*/
 76             n3 = n2.next;
 77             n2.next = n1;
 78             n1 = n2;
 79             n2 = n3;
 80         }
 81         n3 = n1;  //n3->last node
 82         n2 = head;
 83         boolean res = true;
 84         while (n2 != null && n1 != null) {
 85             if (n2.value != n1.value) {
 86 //                return false;   不能return  还要把链表反转回来
 87                 res = false;
 88                 break;
 89             }
 90             n2 = n2.next;
 91             n1 = n1.next;
 92         }
 93         
 94         n1 = n3.next;
 95         n3.next = null;
 96         while (n1 != null) { // recover list
 97             n2 = n1.next;
 98             n1.next = n3;
 99             n3 = n1;
100             n1 = n2;
101         }
102         return res;
103         
104     }
105     
106     
107     public static void printLinkedList(Node node) {
108         System.out.print("Linked List: ");
109         while (node != null) {
110             System.out.print(node.value + " ");
111             node = node.next;
112         }
113         System.out.println();
114     }
115 
116     
117     public static void main(String[] args) {
118         Node head = null;
119         head = new Node(1);
120         head.next = new Node(2);
121         head.next.next = new Node(2);
122         head.next.next.next = new Node(2);
123 //        head.next.next.next.next = new Node(1);
124         printLinkedList(head);
125         System.out.print(isPalindrome1(head) + " | ");
126         System.out.print(isPalindrome2(head) + " | ");
127         System.out.print(isPalindrome3(head) + " | ");
128         printLinkedList(head);
129         System.out.println("=========================");
130 
131     }
132 }

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