Reverse Nodes in k-Group

描述
Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.
You may not alter the values in the nodes, only nodes itself may be changed.
Only constant memory is allowed.
For example, Given this linked list: 1->2->3->4->5
For k = 2, you should return: 2->1->4->3->5
For k = 3, you should return: 3->2->1->4->5
分析
分塊翻轉鏈表，將每 K 個節點作爲一個鏈表翻轉，翻轉之後原來塊的頭結點作爲尾節點指向鏈表中的下一個節點。

// Reverse Nodes in k-Group
// 遞歸版，時間複雜度O(n)，空間複雜度O(1)
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        if (head == null || head.next == null || k < 2)
            return head;

        ListNode next_group = head;
        for (int i = 0; i < k; ++i) {
            if (next_group != null)
                next_group = next_group.next;
            else
                return head;
        }
        // next_group is the head of next group
        // new_next_group is the new head of next group after reversion
        ListNode new_next_group = reverseKGroup(next_group, k);
        ListNode prev = null, cur = head;
        while (cur != next_group) {
            ListNode next = cur.next;
            cur.next = prev != null ? prev : new_next_group;
            prev = cur;
            cur = next;
        }
        return prev; // prev will be the new head of this group
    }
}

// Reverse Nodes in k-Group
// 迭代版，時間複雜度O(n)，空間複雜度O(1)
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        if (head == null || head.next == null || k < 2) return head;
        ListNode dummy = new ListNode(-1);
        dummy.next = head;

        for(ListNode prev = dummy, end = head; end != null; end = prev.next) {
            for (int i = 1; i < k && end != null; i++)
                end = end.next;
            if (end  == null) break;  // 不足 k 個

            prev = reverse(prev, prev.next, end);
        }

        return dummy.next;
    }

    // prev 是 first 前一個元素, [begin, end] 閉區間，保證三者都不爲 null
    // 返回反轉後的倒數第1個元素
    ListNode reverse(ListNode prev, ListNode begin, ListNode end) {
        ListNode end_next = end.next;
        for (ListNode p = begin, cur = p.next, next = cur.next;
                cur != end_next;
                p = cur, cur = next, next = next != null ? next.next : null) {
            cur.next = p;
        }
        begin.next = end_next;
        prev.next = end;
        return begin;
    }
};