234. 回文链表

简单

题目描述

解法一：链表值移至数组

class Solution {
    public boolean isPalindrome(ListNode head) {
        List<Integer> vals = new ArrayList<Integer>();

        // 将链表的值复制到数组中
        ListNode currentNode = head;
        while (currentNode != null) {
            vals.add(currentNode.val);
            currentNode = currentNode.next;
        }

        // 使用双指针判断是否回文
        int front = 0;
        int back = vals.size() - 1;
        while (front < back) {
            if (!vals.get(front).equals(vals.get(back))) {
                return false;
            }
            front++;
            back--;
        }
        return true;
    }
}

解法二：快慢指针

链表后半部分可以使用反转链表

Java

class Solution {
    public boolean isPalindrome(ListNode head) {
        ListNode fast = head;
        ListNode slow = head;
        LinkedList<Integer> list = new LinkedList<>();
        while (fast.next != null && fast.next.next != null){
            list.add(slow.val);
            slow = slow.next;
            fast = fast.next.next;
        }
        if(fast.next != null) list.add(slow.val);
        slow = slow.next;
        while (slow != null) {
            if(list.get(list.size() - 1) != slow.val) return false;
            list.remove(list.size() - 1);
            slow = slow.next;
        }
        return true;
    }
}

Python

class Solution:
    # 876. 链表的中间结点
    def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]:
        slow = fast = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        return slow

    # 206. 反转链表
    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        pre, cur = None, head
        while cur:
            nxt = cur.next
            cur.next = pre
            pre = cur
            cur = nxt
        return pre

    def isPalindrome(self, head: Optional[ListNode]) -> bool:
        mid = self.middleNode(head)
        head2 = self.reverseList(mid)
        while head2:
            if head.val != head2.val:  # 不是回文链表
                return False
            head = head.next
            head2 = head2.next
        return True