234. Palindrome Linked List

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?

Solution1  3ms 21.69 and this is O(1)space

public class Solution {
public boolean isPalindrome(ListNode head) {
if(head==null || head.next == null){
return true;
}
ListNode fast = head;
ListNode slow = head;
Stack<Integer> stack = new Stack<Integer>();
while(fast != null && fast.next != null){
stack.push(slow.val);
slow = slow.next;
fast = fast.next.next;
}
if(fast != null){
slow = slow.next;
}
while(slow != null){
int val = stack.pop();
if(slow.val != val){
return false;
}
slow = slow.next;
}
return true;
}
}

solution2 2ms 34.78

find the middle node of the linked list, and reverse the second half.

public class Solution {
public boolean isPalindrome(ListNode head) {
if(head==null || head.next == null){
return true;
}
ListNode fast = head;
ListNode slow = head;
Stack<Integer> stack = new Stack<Integer>();
while(fast != null && fast.next != null){
stack.push(slow.val);
slow = slow.next;
fast = fast.next.next;
}
if(fast != null){// Notice here you cannot use fast.next == null
slow = slow.next;
}

ListNode next = slow.next;
slow.next = null;
ListNode nextNext;
while(next != null){
nextNext = next.next;
next.next = slow;
slow = next;
next = nextNext;
}
while(slow != null){
if(slow.val != head.val){
return false;
}
slow = slow.next;
head = head.next;
}
return true;
}
}