题解 | #链表中的节点每k个一组翻转#

蒙逼解题法：debug一个小时，解出来头也昏了

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param head ListNode类 
     * @param k int整型 
     * @return ListNode类
     */
    public ListNode reverseKGroup (ListNode head, int k) {
        // write code here
        ListNode left = null;
        ListNode right = null;
        ListNode cur = head;
        ListNode count = head;
       // ListNode newHead = null;
        int len = 0;
        //求链表长度
        while(count != null){
            len++;
            count = count.next;
        }
        if(len == 0) return null;
        if(len == 1) return head;
        if(k == 1) return head;
        //将长链表分割成大小为k的小链表遍历
        for(int j=1;j<=len/k;j++){
            ListNode temp = left;
            left = cur;
            //遍历k，找到待分割部分的left和right
            for(int i=1;i<=k;i++){
                //if(cur == null) return head;
                cur = cur.next;
                //if(k == 1){right = cur; cur= cur.next; break;}
                if(i == k-1) right = cur;
               
            }
            //斩断链表
            right.next = null;
            //反转部分链表
            reverse(left);
            left.next = cur;
            if(j == 1){
                head = right;
            }else{
                temp.next = right;
            }
        }
        
        return head;
    
    }
    //反转截取到的链表
    public void reverse(ListNode head){
        ListNode pre = null;
        ListNode cur2 = head;
        while(cur2 != null){
            ListNode cur_next = cur2.next;
            cur2.next = pre;
            pre = cur2;
            cur2 = cur_next;
            
        }
        
    }
    
}