使用 O(1) 额外空间查找链表中最长回文列表长度的Java程序

给定一个链表，找出该链表中存在的最长回文链表的长度。
例子：

Input  : List = 2->3->7->3->2->12->24
Output : 5
The longest palindrome list is 2->3->7->3->2

Input  : List = 12->4->4->3->14
Output : 2
The longest palindrome list is 4->4

一个简单的解决方案可能是将链表内容复制到数组中，然后在数组中找到最长的回文子数组，但这种解决方案是不允许的，因为它需要额外的空间。
这个想法是基于迭代链表逆过程。我们遍历给定的链表，并从左侧开始一个接一个地反转链表的每个前缀。反转前缀后，我们找到从反转前缀开始的最长公共列表和反转前缀之后的列表。
下面是上述思想的实现。

Java

// Java program to find longest palindrome
// sublist in a list in O(1) time.
class GfG
{
 
//structure of the linked list
static class Node
{
    int data;
    Node next;
}
 
// function for counting the common elements
static int countCommon(Node a, Node b)
{
    int count = 0;
 
    // loop to count common in the list starting
    // from node a and b
    for (; a != null && b != null;
            a = a.next, b = b.next)
 
        // increment the count for same values
        if (a.data == b.data)
            ++count;
        else
            break;
 
    return count;
}
 
// Returns length of the longest palindrome
// sublist in given list
static int maxPalindrome(Node head)
{
    int result = 0;
    Node prev = null, curr = head;
 
    // loop till the end of the linked list
    while (curr != null)
    {
        // The sublist from head to current
        // reversed.
        Node next = curr.next;
        curr.next = prev;
 
        // check for odd length
        // palindrome by finding
        // longest common list elements
        // beginning from prev and
        // from next (We exclude curr)
        result = Math.max(result,
                    2 * countCommon(prev, next)+1);
 
        // check for even length palindrome
        // by finding longest common list elements
        // beginning from curr and from next
        result = Math.max(result,
                    2*countCommon(curr, next));
 
        // update prev and curr for next iteration
        prev = curr;
        curr = next;
    }
    return result;
}
 
// Utility function to create a new list node
static Node newNode(int key)
{
    Node temp = new Node();
    temp.data = key;
    temp.next = null;
    return temp;
}
 
/* Driver code*/
public static void main(String[] args)
{
    /* Let us create a linked lists to test
    the functions
    Created list is a: 2->4->3->4->2->15 */
    Node head = newNode(2);
    head.next = newNode(4);
    head.next.next = newNode(3);
    head.next.next.next = newNode(4);
    head.next.next.next.next = newNode(2);
    head.next.next.next.next.next = newNode(15);
 
    System.out.println(maxPalindrome(head));
}
}
 
// This code is contributed by
// Prerna Saini.

输出 ：

5

时间复杂度： O(n ² )
请注意，上面的代码修改了给定的链表，如果不允许修改链表，则可能无法正常工作。然而，我们终于可以再做一次逆向来获取原始列表。有关更多详细信息，请参阅有关使用 O(1) 额外空间的链表中最长回文列表长度的完整文章！