某些测试用例未能逆转双向链接列表

Question

My implementation for a doubly linked list is listed down below, but for some reason I am not passing one test case. The reverse function only gives us the head of a doubly linked list, and does NOT give us the tail. Is there a certain edge case that I could be missing? ` // Complete the reverse function below.

/*
 * For your reference:
 *
 * DoublyLinkedListNode {
 *     int data;
 *     DoublyLinkedListNode next;
 *     DoublyLinkedListNode prev;
 * }
 *
 */
static DoublyLinkedListNode reverse(DoublyLinkedListNode head) {
    // If the linked list is empty, return null
    if (head == null) {
        return null;
    }

    // If the linked list has only one element, return head becaue reverse of one ele is               itself
    else if (head.next == null) {
        return null;
    }

    // Otherwise reverse
    else {
        DoublyLinkedListNode current = head.next;
        head.next = null; 

        while (current != null) {
            DoublyLinkedListNode nextCurr = current.next; 
            DoublyLinkedListNode prevCurr = current.prev; 
            current.next = prevCurr; 
            current.prev = nextCurr; 
            head = current; 
            current = nextCurr; 
        }

        return head;
    }

}

Answer 1

These logics are wrong:

// If the linked list has only one element, return head becaue reverse of one elem is itself
 else if (head.next == null) {
        return null;    //return head instead (unchanged).
  }

Start with head :

DoublyLinkedListNode current = head.next;   // <<<< current = head
head.next = null;  // <<<< comment out this line

In while loop :

No need to update head each time. Update it with the current at the end of the loop.

Answer 2

I've removed the unnecessary and incorrect logic and variables.

public static DoublyLinkedListNode reverse(DoublyLinkedListNode head) {
    while (head != null) {
        DoublyLinkedListNode nextCurr = head.next;
        head.next = head.prev;
        head.prev = nextCurr;
        if (nextCurr == null) {
            break;
        }
        head = nextCurr;
    }
    return head;
}