leetcode 25. K 个一组翻转链表

题目

class Solution {
    public ListNode caltail(ListNode pretail,ListNode head,int k){//上一组的尾巴,当前组的头 返回当前组的尾巴
        //反转完之后 将pretail的next指向反转后的当前组的头，反转后当前组的尾巴指向原来头指向的next
        ListNode curhead=head,curtail=null,curnext=null;//当前组未反转前的head,tail,next;
        int i=1;
        ListNode now=head,tmp=now.next,tmpnext=(tmp!=null)?tmp.next:null;
        for(;i<k&&tmp!=null;++i){
            if(i==k-1){curtail=tmp;curnext=tmpnext;}
            tmp.next=now;now=tmp;tmp=tmpnext;//反转
            tmpnext=(tmp!=null)?tmp.next:null;
        }
        if(i!=k){//说明最后组不够k个 将他们重新反转过来
            curtail=now;curnext=null;//更新curtail
            //将其反转过来
            tmp=now.next;tmpnext=(tmp!=head&&tmp!=null)?tmp.next:null;
            //这里直接 tmpnext=(tmp!=null)?tmp.next:null; 就可以啦
            while(now!=head){
                tmp.next=now;now=tmp;tmp=tmpnext;
                tmpnext=(tmp!=null)?tmp.next:null;
            }
            pretail.next=curhead;
            curtail.next=curnext;
            return curtail;
        }
        pretail.next=curtail;
        curhead.next=curnext;
        return curhead;
    }
    public ListNode reverseKGroup(ListNode sb, int k) {
        if(k==1) return sb;
        ListNode head=new ListNode();
        head.next=sb;
        ListNode now=head;
        while(now.next!=null){now=caltail(now,now.next,k);}
        return head.next;
    }
}