Reverse Nodes In K Group

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

	public ListNode reverseKGroup(ListNode head, int k) {
		// Start typing your Java solution below
		// DO NOT write main() function
		if(head == null || k == 1)
			return head;
		ListNode cur = head;
		int counter = 1;
		while(counter < k && cur.next != null){
			cur = cur.next;
			counter++;
		}
		if(counter < k)
			return head;
		cur = head;
		ListNode next = head.next;
		ListNode temp;
		counter = 1;
		while(counter < k){
			temp = next.next;
			next.next = cur;
			cur = next;
			next = temp;
			counter++;// after one iteration: cur->head; next->remains...
		}
		head.next = reverseKGroup(next, k);
		return cur;
	}

函数运行完一次以后，cur变成实际上的head，而head因为没有被操作过，实际已经成为当前K个Nodes的尾巴。所以才会有head.next=....