Reorder List

描述：
Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You may not modify the values in the list’s nodes, only nodes itself may be changed.

Example 1:

Given 1->2->3->4, reorder it to 1->4->2->3.
Example 2:

Given 1->2->3->4->5, reorder it to 1->5->2->4->3.

分析：
題目規定要in-place，也就是說只能使用O(1)的空間。
可以找到中間節點，斷開，把後半截單鏈表reverse一下，再合併兩個單鏈表。

// Reorder List
// 時間複雜度O(n)，空間複雜度O(1)
class Solution {
    public void reorderList(ListNode head) {
        if (head == null || head.next == null) return;

        ListNode slow = head, fast = head, prev = null;
        while (fast != null && fast.next != null) {
            prev = slow;
            slow = slow.next;
            fast = fast.next.next;
        }
        prev.next = null; // cut at middle

        slow = reverse(slow);

        // merge two lists
        ListNode curr = head;
        while (curr.next != null) {
            ListNode tmp = curr.next;
            curr.next = slow;
            slow = slow.next;
            curr.next.next = tmp;
            curr = tmp;
        }
        curr.next = slow;
    }

    ListNode reverse(ListNode head) {
        if (head == null || head.next == null) return head;

        ListNode prev = head;
        for (ListNode curr = head.next, next = curr.next; curr != null;
            prev = curr, curr = next, next = next != null ? next.next : null) {
                curr.next = prev;
        }
        head.next = null;
        return prev;
    }
}