【重点】234. 回文链表

题目

Python

# 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: Optional[ListNode]) -> bool:
        res = list()
        cur = head
        while cur:
            res.append(cur.val)
            cur = cur.next
        
        i, j = 0, len(res)-1
        while i <= j:
            if res[i] != res[j]:
                return False
            i += 1
            j -= 1
        
        return True

Java

法1：递归，O(N) + O(N)

class Solution {
    ListNode left; // 标记左侧TreeNode
    public boolean isPalindrome(ListNode head) {
        if (head == null) {
            return true;
        }
        left = head;
        return judge(head);
    }

    public boolean judge(ListNode root) {
        if (root == null) {
            return true;
        }
        boolean res = judge(root.next);
        res = res && (left.val == root.val);
        left = left.next;
        return res;
    }
}

如果把空间复杂度改为O(1)，则会非常复杂。