面试题 02.06. 回文链表

English Version

题目描述

编写一个函数，检查输入的链表是否是回文的。

 

示例 1：

输入： 1->2
输出： false

示例 2：

输入： 1->2->2->1
输出： true

 

进阶：
你能否用 O(n) 时间复杂度和 O(1) 空间复杂度解决此题？

解法

先用快慢指针找到链表的中点，接着反转右半部分的链表。然后同时遍历前后两段链表，若前后两段链表节点对应的值不等，说明不是回文链表，否则说明是回文链表。

Python3

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def isPalindrome(self, head: ListNode) -> bool:
        if head is None or head.next is None:
            return True
        slow, fast = head, head.next
        while fast and fast.next:
            slow, fast = slow.next, fast.next.next
        pre, cur = None, slow.next
        while cur:
            t = cur.next
            cur.next = pre
            pre, cur = cur, t
        while pre:
            if pre.val != head.val:
                return False
            pre, head = pre.next, head.next
        return True

Java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public boolean isPalindrome(ListNode head) {
        if (head == null || head.next == null) {
            return true;
        }
        ListNode slow = head;
        ListNode fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode cur = slow.next;
        slow.next = null;
        ListNode pre = null;
        while (cur != null) {
            ListNode t = cur.next;
            cur.next = pre;
            pre = cur;
            cur = t;
        }
        while (pre != null) {
            if (pre.val != head.val) {
                return false;
            }
            pre = pre.next;
            head = head.next;
        }
        return true;
    }
}

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {boolean}
 */
var isPalindrome = function (head) {
    if (!head || !head.next) {
        return true;
    }
    let slow = head;
    let fast = head.next;
    while (fast && fast.next) {
        slow = slow.next;
        fast = fast.next.next;
    }
    let cur = slow.next;
    slow.next = null;
    let pre = null;
    while (cur) {
        let t = cur.next;
        cur.next = pre;
        pre = cur;
        cur = t;
    }
    while (pre) {
        if (pre.val !== head.val) {
            return false;
        }
        pre = pre.next;
        head = head.next;
    }
    return true;
};

C#

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public bool IsPalindrome(ListNode head) {
        if (head == null || head.next == null)
        {
            return true;
        }
        ListNode slow = head;
        ListNode fast = head.next;
        while (fast != null && fast.next != null)
        {
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode cur = slow.next;
        slow.next = null;
        ListNode pre = null;
        while (cur != null)
        {
            ListNode t = cur.next;
            cur.next = pre;
            pre = cur;
            cur = t;
        }
        while (pre != null)
        {
            if (pre.val != head.val)
            {
                return false;
            }
            pre = pre.next;
            head = head.next;
        }
        return true;
    }
}

TypeScript

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function isPalindrome(head: ListNode | null): boolean {
    if (head == null || head.next == null) return true;
    // 快慢指针定位到中点
    let slow: ListNode = head,
        fast: ListNode = head.next;
    while (fast != null && fast.next != null) {
        slow = slow.next;
        fast = fast.next.next;
    }
    // 翻转链表
    let cur: ListNode = slow.next;
    slow.next = null;
    let prev: ListNode = null;
    while (cur != null) {
        let t: ListNode = cur.next;
        cur.next = prev;
        prev = cur;
        cur = t;
    }
    // 判断回文
    while (prev != null) {
        if (prev.val != head.val) return false;
        prev = prev.next;
        head = head.next;
    }
    return true;
}

Go

func isPalindrome(head *ListNode) bool {
	if head == nil {
		return true
	}
	m := mid(head)
	temp := reverse(m.Next)
	m.Next = nil
	p, q := head, temp
	res := true
	for p != nil && q != nil {
		if p.Val != q.Val {
			res = false
			break
		}
		p = p.Next
		q = q.Next
	}
	m.Next = reverse(temp)
	return res
}

func mid(head *ListNode) *ListNode {
	slow, fast := head, head.Next
	for fast != nil && fast.Next != nil {
		slow = slow.Next
		fast = fast.Next.Next
	}
	return slow
}

func reverse(head *ListNode) *ListNode {
	var prev *ListNode = nil
	for head != nil {
		temp := head.Next
		head.Next = prev
		prev = head
		head = temp
	}
	return prev
}

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