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?

一刷
题解：如果不修改input（即reverse它的下一半的话）是不可能达到O(n) time and O(1) space complexity的。

public boolean isPalindrome(ListNode head) {
    ListNode fast = head, slow = head;
    while (fast != null && fast.next != null) {
        fast = fast.next.next;
        slow = slow.next;
    }
    if (fast != null) { // odd nodes: let right half smaller
        slow = slow.next;
    }
    slow = reverse(slow);
    fast = head;
    
    while (slow != null) {
        if (fast.val != slow.val) {
            return false;
        }
        fast = fast.next;
        slow = slow.next;
    }
    return true;
}

public ListNode reverse(ListNode head) {
    ListNode prev = null;
    while (head != null) {
        ListNode next = head.next;
        head.next = prev;
        prev = head;
        head = next;
    }
    return prev;
}

二刷
同上

/**
 * 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) {
        //reverse the half
        
        //find the medium
        ListNode fast = head, slow = head;
        while(fast!=null && fast.next!=null){
            fast = fast.next.next;
            slow = slow.next;
        }
        //len is odd, from slow.next
        if(fast != null){//odd
            slow = slow.next;
        }
        
        slow = reverse(slow);
        
        fast = head;
        while(slow != null){
            if(slow.val!=fast.val) return false;
            slow = slow.next;
            fast = fast.next;
        }
        return true; 
    }
    
    public ListNode reverse(ListNode head){
        ListNode prev = null;
        while(head!=null){
            ListNode next = head.next;
            head.next = prev;
            prev = head;
            head = next;
        }
        return prev;
    }
}