234. 回文链表

题目描述

给你一个单链表的头节点 head ，请你判断该链表是否为回文链表。如果是，返回 true ；否则，返回 false 。

 

示例 1：

输入：head = [1,2,2,1]
输出：true

示例 2：

输入：head = [1,2]
输出：false

 

提示：

• 链表中节点数目在范围[1, 105] 内
• 0 <= Node.val <= 9

 

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？

方法一：快慢指针

我们可以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。

时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为链表的长度。

Java

class Solution {
    public boolean isPalindrome(ListNode head) {
        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;
    }
}

C++

class Solution {
public:
    bool isPalindrome(ListNode* head) {
        ListNode* slow = head;
        ListNode* fast = head->next;
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        ListNode* pre = nullptr;
        ListNode* cur = slow->next;
        while (cur) {
            ListNode* 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;
    }
};

TypeScript

function isPalindrome(head: ListNode | null): boolean {
    let slow: ListNode = head,
        fast: ListNode = head.next;
    while (fast != null && fast.next != null) {
        slow = slow.next;
        fast = fast.next.next;
    }
    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;
    }
    while (prev != null) {
        if (prev.val != head.val) return false;
        prev = prev.next;
        head = head.next;
    }
    return true;
}

Python

class Solution:
    def isPalindrome(self, head: Optional[ListNode]) -> bool:
        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