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

首先判断是否需要反转元素。

如果需要，采用头插法重新插入元素。

	    public ListNode reverseKGroup(ListNode head, int k) {
	        if(head == null || head.next == null || k == 1){
	            return head;
	        }
	        ListNode first = new ListNode(0);
	        first.next = head;
	        ListNode start = first;
	        ListNode end = first;
	        ListNode temp = head;
	        int i = 0;
	        int flag = 0 ;
	        while(flag == 0){
	            
	            if(i == 0){
	                ListNode list = temp;
	                int j = 1;
	                while(list != null && j < k){
	                    list = list.next;
	                    j ++;
	                }
	                if(j == k && list != null){
	                    end = temp;
	                    temp = temp.next;
	                    i ++;
	                }else{
	                    flag = 1;
	                }

	            }else{
	                end.next = temp.next;
	                temp.next = start.next;
	                start.next = temp;
	                if(i == k -1){
	                    i = 0;
	                    start = end;
	                }else{
	                    i ++;
	                }
	                temp = end.next;
	            }
	        }
	        return first.next;
	    }