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?
先找到链表中点，将第二部分反转，然后比较两部分链表的值。
/**
* 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;
}
// Find the mid element
ListNode fast = head;
ListNode slow = head;
while(fast.next != null && fast.next.next != null) {
slow = slow.next;
fast = fast.next.next;
}
ListNode head2 = slow.next;
slow.next = null;
// Reverse second half
ListNode p1 = head2;
ListNode p2 = head2.next;
while(p2 != null && p1 != null) {
ListNode temp = p2.next;
p2.next = p1;
p1 = p2;
p2 = temp;
}
head2.next = null;
// Compare two parts
ListNode p = p2 == null? p1: p2;
ListNode q = head;
while (p != null) {
if (p.val != q.val) {
return false;
}
p = p.next;
q = q.next;
}
return true;
}
}
)
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