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

解題思想：

首先題目的意思是說給定一個連結串列和一個數k，將連結串列以k個結點為一組進行反轉，如果不夠k個結點則不進行反轉。主要分為以下幾步來進行解題：

1.  將連結串列以k個結點為一組進行拆分，如果夠k個結點則進行反轉

2.  如果不夠k個結點則不進行反轉

3.  拼接階段，將反轉後的一小段連結串列插入到已經操作過的連結串列中間，如果不夠k個結點則不操作。

具體見下面的Java程式碼（有詳細的註釋）：

public class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
		if( head == null || head.next == null ){
			return head;
		}
		
		ListNode p = head;
		ListNode pre = null;        //pre 記錄上一個半段連結串列的尾結點
		boolean flag = true;        //為是否是頭結點做標記
		while( p != null ){
			ListNode tail = p;     //當前是頭，但是反過來就不是頭了，成了尾部
			int i;
			for( i = 1; i < k && p != null; ++i ){
				p = p.next;
			}
			
			ListNode nextList = null;
			if( i >= k && p != null ){
				nextList = p.next;
				p.next = null;   //將當前連結串列的最後一個節點next置為null
				ListNode tempHead = reverse(tail);
				if( flag == true ){  //如果頭，進行處理
					head = tempHead;
					flag = false;
					tail.next = nextList;
				}else{             //如果不是頭的話
					tail.next = nextList;
					pre.next = tempHead;
				}
			}
			pre = tail;     //記錄上一段連結串列的尾結點
			p = nextList;
		}
		
		return head;
    }
	
	public ListNode reverse( ListNode head ){
		if( head == null || head.next == null ){
			return head;
		}
		ListNode p = head, q = head.next;
		head.next = null;
		while( q != null ){
			ListNode r = q.next;
			q.next = p;
			
			p = q;
			q = r;
		}
		head = p;
		return head;
	}
}