请判断一个链表是否为回文链表。
示例 1:
输入: 1->2 输出: false
示例 2:
输入: 1->2->2->1 输出: true
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?
思路:利用快慢指针找到中间节点,当快指针走到末尾时, 慢指针指向中间节点;交中间节点之后的节点进行链表反转; 设定指针p1从head开始, p2从中间节点下一节点开始, 逐个比较,如果有不相等则不是回文,如果都相等,则是回文。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
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) {
slow = slow.next;
fast = fast.next.next;
}
ListNode reverseHead = reverseList(slow.next);
while (head != null && reverseHead != null) {
if (head.val != reverseHead.val)
return false;
head = head.next;
reverseHead = reverseHead.next;
}
return true;
}
public ListNode reverseList(ListNode head) {
if (head == null || head.next == null)
return head;
ListNode p = null;
ListNode q;
while (head != null) {
q = head.next;
head.next = p;
p = head;
head = q;
}
return p;
}
}