Follow up:
Could you do it in O(n) time and O(1) space?
判断一个单向链表是否为回文链表,要求时间复杂度为O(n),空间复杂度为O(1)。我们可以先用快慢指针找到中间的节点,然后将后半部分链表反转,然后以前半部分的链表进行比较,如果相同就返回true,如果不同就返回false。代码如下:
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public boolean isPalindrome(ListNode head) { if(head == null || head.next == null) return true; ListNode fast = head; ListNode slow = head; while(fast.next != null && fast.next.next != null) { fast = fast.next.next; slow = slow.next; } ListNode newHead = slow.next; slow.next = null; newHead = invert(newHead); while(newHead != null) { if(head.val != newHead.val) return false; head = head.next; newHead = newHead.next; } return true; } public ListNode invert(ListNode head) { ListNode pre = null; ListNode tem = null; while(head != null) { tem = head.next; head.next = pre; pre = head; head = tem; } return pre; } }