判断一个链表是否为回文结构

1.题目描述 

2.解法一：时间O(N) ,空间O(N).

3.解法二：时间O(N) ,空间O(N/2).

4.解法三：时间O(N) ,空间O(1).

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1.题目描述 

【 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)。
 

 2.解法一：时间O(N) ,空间O(N).

public class IsPalindromeList {

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

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

	// need n extra space
	public static boolean isPalindrome1(Node head) {
		Stack<Node> stack = new Stack<Node>();
		Node cur = head;
		while (cur != null) { // 全部入栈，然后出栈比较
			stack.push(cur);
			cur = cur.next;
		}
		while (head != null) {
			if (head.value != stack.pop().value) {
				return false;
			}
			head = head.next;
		}
		return true;
	}
}

  3.解法二：时间O(N) ,空间O(N/2).

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) {  // 仅后半段入栈
			stack.push(right);
			right = right.next;
		}
		while (!stack.isEmpty()) {
			if (head.value != stack.pop().value) {
				return false;
			}
			head = head.next;
		}
		return true;
	}

  4.解法三：时间O(N) ,空间O(1).

	// need O(1) extra space
    // 思路：将链表的后半段逆序，如1->2->3->2->1  =>  1->2->3<-2<-1,(然后3指向null)
	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;
	}