解法一:递归
如果不考虑递归的栈开销的话,满足空间复杂度的要求。
/*** Definition for singly-linked list.* public class ListNode {* int val;* ListNode next;* ListNode(int x) { val = x; }* }*/class Solution {ListNode cur;public boolean isPalindrome(ListNode head) {cur = head;return dfs(head);}private boolean dfs(ListNode p) {if (p == null) {return true;}boolean ans = dfs(p.next) && (p.val == cur.val);cur = cur.next;return ans;}}
