234. Palindrome Linked List

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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;    }}

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