Reverse Nodes in k-Group [逆转K-Group链表]

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 class Solution {
	public ListNode reverseKGroup(ListNode head, int k) {
		if(head==null){
			return null;
		}
		int len=1;
		int i=0;
		int group=0;
		int r=0;
		ListNode p=head;
		//获取链表长度
		while(p.next!=null){
			len++;
			p=p.next;
		}
		if(head.next==null || k>len){
			return head;
		}

		//计算整组总数
		group=len/k;
		r=len%k;

		//倒序节点
//		ListNode fNode=null,pNode=head,next;
//		while(pNode!=null){
//			next=pNode.next;
//			pNode.next=fNode;
//			fNode=pNode;//当前节点下一轮就是别人的next了
//			pNode=next;
//		}
//		head=fNode;
		ListNode pNode=head,rNode=null,nNode=null;
		//用于记录当前位置
		int index=0;
		for(i=0;i<group;i++){
			index=i*k;
			if(index==len-r-1){
				break;
			}
//			int startIndex=i*k;
			if(i==0){
				rNode=Solution.reversePart_(pNode, k);
			}else{
				nNode=pNode.next;
				Solution.reversePart(pNode, k);
				pNode=nNode;
			}
		}
		return rNode;
	}

	//获取指定位置的节点
	public static ListNode getListNodeForIndex(ListNode n,int l){
		while(n!=null && l>0){
			n=n.next;
			l--;
		}
		return n;
	}
	//逆转从n.next到指定长度l
	public static ListNode reversePart(ListNode pNode,int l){
		ListNode next=null,fNode=null,fistNode=null,n=pNode.next;
		fistNode=n;
		while(n!=null && l>0){
			next=n.next;
			n.next=fNode;
			fNode=n;
			n=next;
			l--;
		}
		if(n!=null){
			fistNode.next=n;
		}
		pNode.next=fNode;
		return fNode;
	}
	//逆转从n到指定长度l
		public static ListNode reversePart_(ListNode n,int l){
			ListNode next=null,fNode=null,fistNode=null;
			fistNode=n;
			while(n!=null && l>0){
				next=n.next;
				n.next=fNode;
				fNode=n;
				n=next;
				l--;
			}
			if(n!=null){
				fistNode.next=n;
			}
			return fNode;
		} 

}