LeetCode 234 - Palindrome Linked List

Given a singly linked list of characters, write a function that returns true if the given list is palindrome, else false.

METHOD 1 (Use a Stack)
A simple solution is to use a stack of list nodes. This mainly involves three steps.
1) Traverse the given list from head to tail and push every visited node to stack.
2) Traverse the list again. For every visited node, pop a node from stack and compare data of popped node with currently visited node.
3) If all nodes matched, then return true, else false.

Time complexity of above method is O(n), but it requires O(n) extra space. Following methods solve this with constant extra space.

public boolean isListPalindrome(ListNode head) {
	Stack<ListNode> stack = new Stack<>();
	ListNode node = head;
	while(node != null) {
		stack.push(node);
		node = node.next;
	}
	while(head != null) {
		if(head.val != stack.pop().val) return false;
		head = head.next;
	}
	return true;
}

 

METHOD 2 (By reversing the list)
This method takes O(n) time and O(1) extra space.
1) Get the middle of the linked list.
2) Reverse the second half of the linked list.
3) Check if the first half and second half are identical.
4) Construct the original linked list by reversing the second half again and attaching it back to the first half

public boolean isListPalindrome(ListNode head) {
	if(head==null || head.next==null) return true;
	ListNode walker = head, runner = head;
	while(walker!=null && runner!=null) {
		walker = walker.next;
		runner = runner.next;
		if(runner!=null) {
			runner = runner.next;
		}
	}
	ListNode h = reverseLinkedList(walker);
	while(h!=null && head!=null) {
		if(h.val != head.val) return false;
		h = h.next;
		head = head.next;
	}
	return true;
}

public ListNode reverseLinkedList(ListNode head) {
	ListNode dummy = new ListNode(0);
	ListNode pre = dummy;
	while(head != null) {
		ListNode next = head.next;
		head.next = pre.next;
		pre.next = head;
		head = next;
	}
	return dummy.next;
}

重构了下代码：

public boolean isPalindrome(ListNode head) {
    if(head == null || head.next == null) return true;
    ListNode slow = head, fast = head.next;
    while(fast != null && fast.next != null) {
        slow = slow.next;
        fast = fast.next;
        if(fast != null) fast = fast.next;
    }
    ListNode h = slow.next;
    slow.next = null;
    while(h != null) {
        ListNode next = h.next;
        h.next = slow.next;
        slow.next = h;
        h = next;
    }
    ListNode l1 = head, l2 = slow.next;
    while(l2 != null) {
        if(l1.val != l2.val) return false;
        l1 = l1.next;
        l2 = l2.next;
    }
    return true;
}

 

另外，还可以用递归的方法。

递归方法参见：

http://www.geeksforgeeks.org/function-to-check-if-a-singly-linked-list-is-palindrome/