在Java中使用一个只要ArrayList的指数 [英] Using a long as ArrayList index in java

问题描述

我写这篇java程序找到所有的素数高达使用埃拉托色尼的筛NUM，但是当我尝试编译，它说我不能用长var当成数组的索引，它期待一个在它的地方INT变种。但我会用大量的合作，所以我不能使用int类型。我该怎么办？

 进口的java.util。*;
进口的java.lang。*;公共类T3 {
    公共静态无效的主要（字串[] args）{
        长NUM = 100;        //声明列表，并与数字填充它
        ArrayList的＆LT;龙＆GT; numlist =新的ArrayList＆LT;龙＆GT;（）;
        为（长x = 2; X＆LT;民; X ++）{
            numlist.add（新龙（X））;
        }        //筛或埃拉托色尼
        为（长x = 0; X＆LT;的Math.sqrt（NUM）; X ++）{
            为（长Y = X + 1; Y＆LT; numlist.size（）; Y ++）{
                如果（numlist [Y]％numlist [X] == 0）{
                    numlist.remove（Y）;
                }
            }
        }        //打印列表
        对于（Object项：numlist）{
            的System.out.println（（长）项）;
        }
    }
}
 

解决方案

我不知道为什么你的code将汇编开始。

你不应该使用[]数组中的列表访问的成员。一个ArrayList的仅仅是在内部存储在数组中的列表。你必须使用列表get操作（这将仍然是O（1））。写作numlist [指数]是指你在numlist对象的数组。你不能覆盖的[]操作在C ++中。

在此外，int是用Java 32位。其长度大于数组2 ^ 32（所以你需要长指数）是不可能的，我甚至不知道在规范允许的。

I am writing this java program to find all the prime numbers up to num using the Sieve of Eratosthenes, but when I try to compile, it says I can't use a long var as an array index, and it expects an int var in its place. But I'll be working with large numbers, so I can't use int. What can I do?

import java.util.*;
import java.lang.*;

public class t3{
    public static void main(String[] args){
        long num = 100;

        //declaring list and filling it with numbers
        ArrayList<Long> numlist = new ArrayList<Long>();
        for(long x=2 ; x<num ; x++){
            numlist.add(new Long(x));
        }

        //sieve or eratosthenes
        for(long x=0 ; x<Math.sqrt(num) ; x++){
            for(long y=x+1 ; y<numlist.size() ; y++){
                if(numlist[y]%numlist[x] == 0){
                    numlist.remove(y);
                }
            }
        }

        //print list
        for(Object item : numlist){
            System.out.println((Long)item);
        }
    }
}

解决方案

I'm not sure why your code would compile to begin with.

You're not supposed to use [] in an array list to access members. An arraylist is merely a list that is internally stored in an array. You have to use the list get operation (which would still be O(1)). Writing numlist[index] means that you have an array of objects in numlist. You cannot override the [] operation as in C++.

In addition, an int is 32 bits in Java. Having an array of length greater than 2^32 (so you would need long indices) is unlikely and I'm not even sure the specification allows it.

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