将单链表的每k个节点之间逆序

import java.util.Stack;//将单链表的每k个节点之间逆序public class ReverseKOfList{        //单链表节点的定义    public static class Node{      public int value;      public Node next;      public Node(int data)      {      this.value=data;      }    }    //(方法一、反转部分链表法)每k个节点之间逆序    public static Node ReverseK(Node head,int k)    {        if(head==null||k<=0)        {        return head;        }        if(k==1)        {        return head;        }        int n=0; //记录链经历的节点数        Node p=head; //记录链表的头节点        Node q=head;        int leng=0;        while(q!=null)        {            ++leng;            q=q.next;        }        //n=(int)Math.floor(leng/k);        n=leng/k;     //记录反转的次数        //System.out.println(n);        int from=1;        int to=k;        while(n>0)        {             Node mode=reverseSublist(head,from,to); //记录每次变换后的链表头节点          //PrintList(mode);          head=mode;          from+=k;          to+=k;          n--;        }        return head;    }     //反转部分链表   public static Node reverseSublist(Node head, int from,int to)    {    if(head==null||from <0||to<0)    {    return head;    }        Node p=head;    int leng=0;    Node tPre=null; //获取反转起始节点的前一个节点        Node tPos=null; //获取反转结束节点的后一个节点    //计算链表长度    while(p!=null)     {           ++leng;           tPre=(leng==from-1?p:tPre); //找到反转起始节点的前一个节点           tPos=(leng==to+1?p:tPos);    //找到反转结束节点的后一个节点           p=p.next;    }    if(from>leng||to>leng||from>to)    {    return head;    }    p=tPre==null?head:tPre.next;  //找到要起始反转的节点    Node node2=p.next;            p.next=tPos;  //起始反转的节点指向反转结束节点的后一个节点                 Node next=null;        while(node2!=tPos)        {        next=node2.next;        node2.next=p;        p=node2;        node2=next;        }        if(tPre!=null)        {        tPre.next=p;        return head;        }        return p;               }      //(方法二、利用栈结构)每k个节点之间逆序    public static Node ReverseK2(Node head, int K) {if (K < 2) {return head;}Stack<Node> stack = new Stack<Node>();Node newHead = head; //记录新的头结点Node cur = head;Node pre = null;Node next = null;while (cur != null) {next = cur.next;stack.push(cur);if (stack.size() == K) {pre = resign1(stack, pre, next);newHead = newHead == head ? cur : newHead;}cur = next;}return newHead;}    //调整k个节点逆序public static Node resign1(Stack<Node> stack, Node left, Node right) {Node cur = stack.pop();if (left != null) {left.next = cur;}Node next = null;while (!stack.isEmpty()) {next = stack.pop();cur.next = next;cur = next;}cur.next = right;return cur;}   //(方法三、利用变量存储)每k个节点之间逆序public static Node ReverseK3(Node head, int K) {if (K < 2) {return head;}Node cur = head;Node start = null;Node pre = null;Node next = null;int count = 1;while (cur != null) {next = cur.next;if (count == K) {start = pre == null ? head : pre.next;head = pre == null ? cur : head;resign2(pre, start, cur, next);pre = start;count = 0;}count++;cur = next;}return head;}   //调整k个节点逆序public static void resign2(Node left, Node start, Node end, Node right) {Node pre = start;Node cur = start.next;Node next = null;//链表的逆转while (cur != right) {next = cur.next;cur.next = pre;pre = cur;cur = next;}if (left != null) {left.next = end;}start.next = right;}    //打印链表    public static void PrintList(Node head)    {    while(head!=null)    {    System.out.print(head.value+" ");    head=head.next;    }    System.out.println();    }public static void main(String[]args){Node node=new Node(1);node.next=new Node(2);node.next.next=new Node(3);node.next.next.next=new Node(4);node.next.next.next.next=new Node(5);node.next.next.next.next.next=new Node(6);node.next.next.next.next.next.next=new Node(7);node.next.next.next.next.next.next.next=new Node(8);PrintList(node);//Node mode=ReverseK(node,3); //时间、空间复杂度大//Node mode=ReverseK2(node,3);//时间O(n),空间O(k)Node mode=ReverseK3(node,3);  //时间O(n),空间O(1)        PrintList(mode);}}