题解 | #链表中的节点每k个一组翻转#

分段头插法：new一个辅助链表头出来---newhead

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
        int size = 0;
        ListNode p = head;
        while(p!=null){
            size++;
            p = p.next;
        }
        if(head == null || size < k) return head;

        int n = size/k;//一共有多少组链表待翻转
        ListNode newhead = new ListNode(-1);
        newhead.next = head;
        ListNode root = newhead;
        for(int i = 0; i < n ;i++){
            int j = 0;
           while( j < k-1){
               //一个子链表的翻转，头插法
               p  = head.next;
               head.next = p.next;
               p.next = newhead.next;
               newhead.next = p;
               j++;
           }
            //下一个子链表
           newhead = head;
           head = head.next;
        }
        return root.next;
    }
}