reverse-nodes-in-k-group

题目描述

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
我的代码如下：

import java.util.Scanner;import java.util.Stack;class ListNode{    int val;    ListNode next;    ListNode(int x){        val=x;        next=null;    }}public class ReverseNodes {    public static void main(String[] args) {        // TODO Auto-generated method stub        Scanner sc=new Scanner(System.in);        String[] a=sc.nextLine().split(" ");        int k=sc.nextInt();        int[] b=new int[a.length];        for(int i=0;i<a.length;i++)            b[i]=Integer.parseInt(a[i]);        ListNode head=createList(b);        ListNode headSwaped=reverseKGroup(head,k);        while(headSwaped!=null){            System.out.print(headSwaped.val+" ");            headSwaped=headSwaped.next;        }    }    static ListNode reverseKGroup(ListNode head,int k){        if(head==null||head.next==null||k<2)return head;        ListNode first=new ListNode(0);        ListNode pre=first,tail=first,tmp;        first.next=head;        int count;        while(true){            count=k;            while(count>0&&tail!=null){                count--;                tail=tail.next;            }            if(tail==null)break;           pre.next=tail;           pre=head;            while(head!=tail){                tmp=head;                head=tmp.next;                tmp.next=tail.next;                tail.next=tmp;            }            tail=pre;            head=pre.next;        }        return first.next;    }    static ListNode createList(int[] A){        ListNode head=null;        head=new ListNode(A[0]);        ListNode node=head;        for(int i=1;i<A.length;i++){            node.next=new ListNode(A[i]);            node=node.next;        }        return head;    }}