NC50 链表中的节点每k个一组翻转

解法一，借用栈的后进先出完成每个的翻转，时间On 空间Ok，k为栈的借用容量

    public ListNode reverseKGroup (ListNode head, int k) {
        // write code here
        if(k<=1||head==null)
            return head;
        Stack<ListNode> stack = new Stack<>();
        ListNode dummy = new ListNode(-1);
        dummy.next = null;
        ListNode now = dummy;
        ListNode p = head;
        while(p!=null){
            for(int i =0;p!=null&i<k;i++){
                stack.push(p);
                p=p.next;
            }
            if(stack.size()==k){
                while(!stack.isEmpty()){
                    now.next = stack.pop();
                    now = now.next;
                    now.next = null;
                }
            }
        }
        if(!stack.isEmpty()){
            while(stack.size()>1)
                stack.pop();
            now.next = stack.pop();
        }
        return dummy.next;
    }

解法二：模拟，时间On，空间O1

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 length = 0;
        ListNode p = head;
        while(p!=null){
            length++;
            p = p.next;
        }
        if(k<=1||length<k){
            return head;
        }
        int count = length/k;
        ListNode dummy = new ListNode(-1);
        ListNode now = dummy;
        p = head;
        ListNode q = p;
        for(int i =0;i<count;i++){
            ListNode t = p;
            for(int j =0;j<k;j++){
                q = p.next;
                p.next = now.next;
                now.next = p;
                p = q;
            }
            now = t;
        }
        if(p!=null)
            now.next = p;
        return dummy.next;
    }
}