234. 回文链表

回文链表

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？

题解答案

/**
 * 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.next == null) return true;
        if (head.next.next == null) return head.val == head.next.val;
        ListNode mid = middleNode(head);
        ListNode rHead = reverseList(mid.next);
        ListNode lHead = head;
        while (rHead != null) {
            if (lHead.val != rHead.val) {
                return false;
            }
            lHead = lHead.next;
            rHead = rHead.next;
        }
        return true;
    }

    private ListNode middleNode(ListNode head) {
        ListNode fast = head.next;
        ListNode slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }

    private ListNode reverseList(ListNode head) {
        ListNode newHead = null;
        while (head != null) {
            ListNode tmp = head.next;
            head.next = newHead;
            newHead = head;
            head = tmp;
        }
        return newHead;
    }
}

不破坏链表结构

/**
 * 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.next == null) return true;
        if (head.next.next == null) return head.val == head.next.val;
        ListNode mid = middleNode(head);
        ListNode rHead = reverseList(mid.next);
        ListNode rOldHead = rHead;
        ListNode lHead = head;
        boolean result = true;
        while (rHead != null) {
            if (lHead.val != rHead.val) {
                result = false;
                break;
            }
            lHead = lHead.next;
            rHead = rHead.next;
        }
        reverseList(rOldHead);
        return result;
    }

    private ListNode middleNode(ListNode head) {
        ListNode fast = head.next;
        ListNode slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }

    private ListNode reverseList(ListNode head) {
        ListNode newHead = null;
        while (head != null) {
            ListNode tmp = head.next;
            head.next = newHead;
            newHead = head;
            head = tmp;
        }
        return newHead;
    }
}