牛客题霸--链表中的节点每k个一组翻转--题解

牛客题霸题目链接:链表中的节点每k个一组翻转
https://www.nowcoder.com/practice/b49c3dc907814e9bbfa8437c251b028e?tpId=117&&tqId=34971&rp=1&ru=/ta/job-code-high&qru=/ta/job-code-high/question-ranking
这个题目属于高频面试题，希望大家多多练习，也是入门必备

class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode dummy=new ListNode();
        dummy.next=head;
        //定义两个指针
        ListNode start=dummy;
        ListNode end=dummy;
        while(end!=null){
            for(int i=0;i<k && end!=null;i++){
                end=end.next;
            }
            if(end==null)break;
            ListNode temp1=end.next;
            ListNode temp2=start.next;
            end.next=null;
            //找到尾部，进行反转
            start.next=reverse(temp2);
            temp2.next=temp1;
            start=temp2;
            end=temp2;
        }
        return dummy.next;
    }
    //单个链表反转方法
    public ListNode reverse(ListNode head){
        ListNode pre=null;
        while(head!=null){
            ListNode temp=head.next;
            head.next=pre;
            pre=head;
            head=temp;
        }
        return pre;
    }
}