牛客网 在线编程 回文链表

题目描述

https://www.nowcoder.com/practice/baefd05def524a92bcfa6e1f113ed4f0?tpId=8&&tqId=11006&rp=1&ru=/activity/oj&qru=/ta/cracking-the-coding-interview/question-ranking

请编写一个函数，检查链表是否为回文。

给定一个链表ListNode* pHead，请返回一个bool，代表链表是否为回文。

测试样例：

{1,2,3,2,1}

返回：true

{1,2,3,2,3}

返回：false

老样子还是先挂代码

import java.util.*;

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Palindrome {
    public boolean isPalindrome(ListNode head) {
if (head == null) {
			return false;
		}
		if (head.next == null) {
			return true;
		}
		ListNode fp = head;
		ListNode lp = head;
		while (fp.next != null && fp.next.next != null) {
			fp = fp.next.next;
			lp = lp.next;
		}
		ListNode n1 = lp.next;
		ListNode n2 = n1.next;
		ListNode n3 = null;
		lp.next = null;
		n1.next = null;
		while (n2 != null) { // 反向链接需要三个节点循环
			n3 = n2.next;
			n2.next = n1;
			n1 = n2;
			n2 = n3;
		}
		ListNode tail = n1;
		while (head != null && tail != null) {
			if (head.val == tail.val) {
				head = head.next;
				tail = tail.next;
			} else {

				n2 = n1.next;
				n1.next = null;
				while (n2 != null) {
					n3 = n2.next;
					n2.next = n1;
					n1 = n2;
					n2 = n3;
				}
				lp.next = n1;
				return false;
			}
		}
		n2 = n1.next;
		n1.next = null;
		while (n2 != null) {
			n3 = n2.next;
			n2.next = n1;
			n1 = n2;
			n2 = n3;
		}
		lp.next = n1;
		return true;
    }
}

这个代码比较长，但是它的额外空间复杂度为O(1)，用最简单的存储比较方式的话需要空间复杂度O(N)或者O(N/2)

代码的原理主要是用一个快指针fp和一个慢指针lp，当fp移动到最后尾部（链表为奇数）或者尾部前一个（链表为偶数）时，lp会达到左半部分的某位，中间的左边（链表为偶数），或者最中间（链表为奇数）

12345677654321对于这个链表,fp会停在右2,lp会停在左7

1234567654321,fp会停在最后1，lp会停在7

然后将右边的链表结构改成逆序，生成两个新的链表，尾节点都指向null，分别遍历比较就好了。

下面附上左神算法课中的三种回文链表的判断方式，空间复杂度分别为O（N），O(N/2),O(1)

import java.util.*;

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
public class Palindrome {
    public boolean isPalindrome(ListNode head) {
if (head == null) {
			return false;
		}
		if (head.next == null) {
			return true;
		}
		ListNode fp = head;
		ListNode lp = head;
		while (fp.next != null && fp.next.next != null) {
			fp = fp.next.next;
			lp = lp.next;
		}
		ListNode n1 = lp.next;
		ListNode n2 = n1.next;
		ListNode n3 = null;
		lp.next = null;
		n1.next = null;
		while (n2 != null) { // 反向链接需要三个节点循环
			n3 = n2.next;
			n2.next = n1;
			n1 = n2;
			n2 = n3;
		}
		ListNode tail = n1;
		while (head != null && tail != null) {
			if (head.val == tail.val) {
				head = head.next;
				tail = tail.next;
			} else {

				n2 = n1.next;
				n1.next = null;
				while (n2 != null) {
					n3 = n2.next;
					n2.next = n1;
					n1 = n2;
					n2 = n3;
				}
				lp.next = n1;
				return false;
			}
		}
		n2 = n1.next;
		n1.next = null;
		while (n2 != null) {
			n3 = n2.next;
			n2.next = n1;
			n1 = n2;
			n2 = n3;
		}
		lp.next = n1;
		return true;
    }
}