题解 | #链表中的节点每k个一组翻转#

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
        // 先进行分组的倒转
        ListNode ret = new ListNode(-1);
        ListNode pre = ret;
        ListNode preNode = pre;
        int i = 0;
        while(head != null){
            ListNode temp = head;
            head = head.next;
            temp.next = pre.next;
            pre.next = temp;
            i++;
            if(i == k){
                while(pre.next != null){
                    pre = pre.next;
                    preNode = pre;
                }
                i = 0;
            }
        }
        // 若最后一次不足K,再将最后一组再进行一次反转
        if(i != 0){
            ListNode ret2 = new ListNode(-1);
            ListNode pre2 = ret2;
            head = preNode.next;
            while(head != null){
                ListNode temp = head;
                head = head.next;
                temp.next = pre2.next;
                pre2.next = temp;
            }
            preNode.next = ret2.next;
        }
        return ret.next;
    }
}