Reverse Nodes in k-Group

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For example,
Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        if(head == null || k == 1){
            return head;
        }
        
        int len = 0;
        ListNode p = head;
        while(p != null){
            p= p.next;
            len++;
        }
        
        
        ListNode safe = new ListNode(-1);
        safe.next = head;
        
        ListNode pre = safe, cur = head, post = head.next;
        
        //记录需要翻转几次
        int m = len/k;
        
        for(int i = 0; i < m; i++){
            //跳出 while 之后的新循环开始前 cur 和 post指向同一个Node
            post = cur.next;
            
            // 记录翻转了几个Node了
            int j = 0;
            while(post != null){
                ListNode tmp = post.next;
                post.next = cur;
                cur = post;
                post = tmp;
                j++;
                if(j == k-1){
                    break;
                }
            }
            
            ListNode tmp = pre.next;
            pre.next = cur;
            tmp.next = post;
            pre = tmp;
            cur = pre.next;
        }
        
        return safe.next;
    }
    
}