【Leetcode-Easy-234】Palindrome Linked List

【Leetcode-Easy-234】Palindrome Linked List

题目

Given a singly linked list, determine if it is a palindrome.

Follow up:

Could you do it in O(n) time and O(1) space?

思路

双指针找到链表的中间结点；

反转链表的后半部分；

比较链表的前半部分和“后半部分”。

程序

/**
* 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 != null && fast.next != null){
fast = fast.next.next;
slow = slow.next;
}

// slow 是链表后半截的头指针
if (fast != null) slow = slow.next;

ListNode head1 = head;
ListNode head2 = reverseList(slow); // 反转链表的后半截

while (head1 != null && head2 != null){
if (head1.val != head2.val) return false;

head1 = head1.next;
head2 = head2.next;
}

return true;

}

// 反转链表
private ListNode reverseList(ListNode head){
if (head == null || head.next == null) return head;

ListNode first = head;
ListNode second = head.next;
first.next = null;  // 表示链表末尾
while (second != null){
ListNode tempNode = second.next;
second.next = first;
first = second;
second = tempNode;
}
return first;
}
}