Question

我正在尝试编写一个递归程序，可以打印从文件中取出的整数列表的所有排列，我使用HashMap存储列表，其中键是数字，值是键的次数发生在列表中。代码如下：

import java.util.*;
import java.io.*;

public class perm
{
    public static void permute(HashMap<Integer, Integer> map)
    {
        if(map.isEmpty())
        {
            System.out.println();
        }
        else
        {
            Set<Integer> keys = map.keySet();
            for(int num: keys)
            {
                System.out.print(Integer.toString(num)+",");
                int val = map.get(num) -1;
                if(val == 0)
                {
                    map.remove(num);   
                }
                else
                {
                    map.put(num, val);
                }
                permute(map);
            }
        }
    }
    public static void main(String args []) throws FileNotFoundException
    {
        HashMap <Integer, Integer> map = new HashMap<Integer, Integer>();
        Scanner in = new Scanner(new File(args[0]));
        int num;
        while(in.hasNextInt())
        {
            num = in.nextInt();
            if(map.containsKey(num))
            {
                map.put(num, map.get(num)+1);
            }
            else
            {
                map.put(num,1);
            }
        }
        permute(map);
    }
}

我收到并发错误，但我不明白为什么，错误如下：

Exception in thread "main" java.util.ConcurrentModificationException
    at java.util.HashMap$HashIterator.nextNode(HashMap.java:1429)
    at java.util.HashMap$KeyIterator.next(HashMap.java:1453)
    at perm.permute(perm.java:15)
    at perm.permute(perm.java:27)
    at perm.permute(perm.java:27)
    at perm.permute(perm.java:27)
    at perm.main(perm.java:48)

--------------------基于评论编辑的代码--------------------

import java.util.*;
import java.io.*;

public class perm
{
    public static void permute(HashMap<Integer, Integer> map)
    {
        if(map.isEmpty())
        {
            System.out.println();
        }
        else
        {
            Iterator it = map.entrySet().iterator(); 
            //Set<Integer> keys = map.keySet();
            while(it.hasNext())
            //for(int num: keys)
            {
                Map.Entry pair = (Map.Entry)it.next();
                int num = (int) pair.getKey();
                System.out.print(Integer.toString(num)+",");
                int val = (int) pair.getValue()-1;
                //HashMap<Integer, Integer> newMap = map;
                //int val = map.get(num) -1;
                if(val == 0)
                {
                    //newMap.remove(num);
                    it.remove(); 
                }
                else
                {
                    //newMap.put(num, val);
                    pair.setValue(val);
                }
                permute(map);
            }
        }
    }
    public static void main(String args []) throws FileNotFoundException
    {
        HashMap <Integer, Integer> map = new HashMap<Integer, Integer>();
        Scanner in = new Scanner(new File(args[0]));
        int num;
        while(in.hasNextInt())
        {
            num = in.nextInt();
            if(map.containsKey(num))
            {
                map.put(num, map.get(num)+1);
            }
            else
            {
                map.put(num,1);
            }
        }
        permute(map);
    }
}

同样的错误仍然存在。然后我回到初始方法并存储了我试图迭代的HashMap的临时副本，并且仅修改了临时副本而不是原始副本，同样的错误仍然存在。

----基于答案1的代码编辑------

import java.util.*;
import java.io.*;

public class perm
{
    public static void permute(HashMap<Integer, Integer> map)
    {
        if(map.isEmpty())
        {
            System.out.println();
        }
        else
        {
            //Iterator it = map.entrySet().iterator(); 
            HashMap<Integer, Integer> newMap = map;
            Set<Integer> keys = map.keySet();
            //while(it.hasNext())
            for(int num: keys)
            {
                //Map.Entry pair = (Map.Entry)it.next();
                //int num = (int) pair.getKey();
                System.out.print(Integer.toString(num)+",");
                //int val = (int) pair.getValue()-1;
                //HashMap<Integer, Integer> newMap = map;
                int val = map.get(num) -1;
                if(val == 0)
                {
                    newMap.remove(num);
                    //it.remove(); 
                }
                else
                {
                    newMap.put(num, val);
                    //pair.setValue(val);
                }
                permute(newMap);
            }
        }
    }
    public static void main(String args []) throws FileNotFoundException
    {
        HashMap <Integer, Integer> map = new HashMap<Integer, Integer>();
        Scanner in = new Scanner(new File(args[0]));
        int num;
        while(in.hasNextInt())
        {
            num = in.nextInt();
            if(map.containsKey(num))
            {
                map.put(num, map.get(num)+1);
            }
            else
            {
                map.put(num,1);
            }
        }
        permute(map);
    }
}

Answer 1

修改正在迭代的集合时抛出

java.util.ConcurrentModificationException。在您的情况下，您正在迭代并修改HashMap方法中的permute。

如果没有抛出此异常，那么每次运行程序时，最终都会出现奇怪且看似随机的结果，具体取决于从地图中添加或删除的整数的生成hashCode。

Answer 2

您正在获取ConcurrentModificationException，因为您正在尝试在迭代时修改map。理解它的好链接 http://javarevisited.blogspot.com/2012/02/fail-safe-vs-fail-fast-iterator-in-java.html

java - 使用Hashmap生成排列

2 个答案: