java.lang.OutOfMemoryError:Java堆空间和HashMap

时间:2012-10-07 16:45:26

标签: java recursion hashmap out-of-memory

很抱歉再次发布此代码。以前问题是我遇到了堆栈溢出错误,这是通过使用long而不是int来修复的。但是对于n的大值,我在线程“main”中得到了一个异常java.lang.OutOfMemoryError:Java堆空间。 问题:

Given a positive integer n, prints out the sum of the lengths of the Syracuse 
sequence starting in the range of 1 to n inclusive. So, for example, the call:
lengths(3)
will return the the combined length of the sequences:
1
2 1
3 10 5 16 8 4 2 1 
which is the value: 11. lengths must throw an IllegalArgumentException if 
its input value is less than one.

我的代码:

  import java.util.*;


  public class Test {

HashMap<Long,Integer> syraSumHashTable = new HashMap<Long,Integer>();

public Test(){

}

public int lengths(long n)throws IllegalArgumentException{

    int sum =0;

    if(n < 1){
        throw new IllegalArgumentException("Error!! Invalid Input!");
    }   

    else{

        for(int i=1;i<=n;i++){
            sum+=getStoreValue(i);
        }
        return sum;


    }


}

private int getStoreValue(long index){
    int result = 0;

    if(!syraSumHashTable.containsKey(index)){
        syraSumHashTable.put(index, printSyra(index,1));
    }

    result = (Integer)syraSumHashTable.get(index);

     return result;

}

public static int printSyra(long num, int count) {
    if (num == 1) {
        return count;
    }
    if(num%2==0){

        return printSyra(num/2, ++count);
    }

    else{

        return printSyra((num*3)+1, ++count) ;

    }
}


}

由于我必须添加前面数字的总和,我将在线程“main”java.lang.OutOfMemoryError中结束异常:Java堆空间的n​​值很大。我知道哈希表可以帮助加快计算速度。如果遇到我在使用HashMap之前计算过的元素,我如何确保我的递归方法printSyra可以提前返回值。

驱动程序代码:

public static void main(String[] args) {
    // TODO Auto-generated method stub
    Test t1 = new Test();
    System.out.println(t1.lengths(90090249));

    //System.out.println(t1.lengths(3));
}

1 个答案:

答案 0 :(得分:0)

您需要使用迭代方法而不是递归。递归方法会对线程的堆栈跟踪施加压力。

public static int printSyra(long num, int count) {
    if (num == 1) {
        return count;
    }

    while (true) {
            if (num == 1) break; else if (num%2 == 0) {num /= 2; count++;) else {num = (num*3) + 1; count++;} 
    }
    return count;
}