Reverse LinkedList issue in Java

Question

Reverse second half of linked list

Suppose we have 1-2-3-4

Output: 1-2-4-3

Suppose we have 1-2-3-4-5

Output: 1-2-3-5-4 // however I want the output to be 1-2-5-4-3 in the odd condition, how to modify the code below?

public static ListNode reverseSecondHalfList(ListNode head) {
    if (head == null || head.next == null)      return head;
    ListNode fast = head;
    ListNode slow = head;
    while (fast.next != null && fast.next.next != null) {
        fast = fast.next.next;
        slow = slow.next;
    }
    ListNode pre = slow.next;
    ListNode cur = pre.next;
    while (cur != null) {
        pre.next = cur.next;
        cur.next = slow.next;
        slow.next = cur;
        cur = pre.next;
    }
    return head;
}

My method: Firstly find the start position to be swapped, then swap "pre" node and "cur" each time until cur.next!=null

Answer 1

Try this

    public static ListNode reverseSecondHalfList(ListNode head) {
    if (head == null || head.next == null)      return head;

    // Add one more node before head, it will help us find the node which before mid note.
    ListNode newHead  = new ListNode(0);
    newHead.next= head;
    ListNode fast = newHead;
    ListNode slow = newHead;
    while (fast.next != null && fast.next.next != null) {
        fast = fast.next.next;
        slow = slow.next;
    }
    ListNode pre = slow.next;
    ListNode cur = pre.next;
    while (cur != null) {
        pre.next = cur.next;
        cur.next = slow.next;
        slow.next = cur;
        cur = pre.next;
    }
    return head;
}

And I extract some code from one method make code cleaner

  public static ListNode reverseSecondHalfList2(ListNode head) {
    if (head == null || head.next == null)      return head;

    ListNode preMid = getPreMidListNode(head);
    reverse(preMid);
    return head;
}

private static void reverse(ListNode preMid) {
    ListNode pre = preMid.next;
    ListNode cur = pre.next;
    while (cur != null) {
        pre.next = cur.next;
        cur.next = preMid.next;
        preMid.next = cur;
        cur = pre.next;
    }
}

private static ListNode getPreMidListNode(ListNode head) {
    // Add one more node before head, it will help us find the node which before mid note.
    ListNode newHead  = new ListNode(0);
    newHead.next= head;
    ListNode fast = newHead;
    ListNode slow = newHead;
    while (fast.next != null && fast.next.next != null) {
        fast = fast.next.next;
        slow = slow.next;
    }
    return slow;
}