【牛客题霸题解】 链表中的节点每k个一组翻转

前置题目：
NC78反转链表
NC21链表内指定区间反转
这道题每次先尝试往后走k步，如果不能走到k步的话就说明剩下的节点不足k个了，直接返回。
如果能走k步，就说明这一段的区间可以反转，其实也就转化为了链表内指定区间反转。

下面是图示
确定要反转的区间

反转

确定要反转的区间

反转

没办法往后走k步了， 程序结束

c++

class Solution {
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode * front = new ListNode(-1);
        front->next = head;
        ListNode * pre = front;
        ListNode * now = head;
        ListNode * next = NULL;
        ListNode * rStart = pre;
        while(now!=NULL)
        {
            ListNode * rStart = pre;
            ListNode * temp = rStart;
            for(int i = 0 ; i < k ; i++)//先走k步看一下
            {
                temp = temp->next;
                if(temp==NULL){return front->next;}
            }
            rStart->next = temp;
            ListNode *tpre = now;
            pre = temp->next;
            for(int i = 0 ; i < k ; i++)
            {
                next = now->next;
                now->next = pre;
                pre = now;
                now = next;
            }
            pre = tpre;
        }
        return front->next;
    }
};

java

import java.util.*;
public class Solution {
    public ListNode reverseKGroup (ListNode head, int k) {
        ListNode front = new ListNode(-1);
        front.next = head;
        ListNode pre = front;
        ListNode now = head;
        ListNode next = null;

        while(now!=null)
        {
            ListNode rStart = pre;
            ListNode temp = rStart;
            for(int i = 0 ; i < k ; i++)//先走k步看一下
            {
                temp = temp.next;
                if(temp==null){return front.next;}
            }
            rStart.next = temp;
            ListNode tpre = now;
            pre = temp.next;
            for(int i = 0 ; i < k ; i++)
            {
                next = now.next;
                now.next = pre;
                pre = now;
                now = next;
            }
            pre = tpre;
        }
        return front.next;
    }
}

python

class Solution:
    def reverseKGroup(self , head , k ):
        front =ListNode(-1)
        front.next = head
        pre = front
        now = head
        nexts = None

        while now!=None:
            rStart = pre
            temp = rStart
            for i in range(k):
                temp = temp.next
                if temp==None:
                    return front.next
            rStart.next = temp
            tpre = now
            pre = temp.next
            for i in range(k):
                nexts = now.next
                now.next = pre
                pre = now
                now = nexts
            pre = tpre
        return front.next