-2

根本没有打印与上面的行相同的输出，我不知道为什么会这样，我注意到它正在从末尾向后打印最后 N 个数字，无论我输入到参数中的任何内容，它都会在一秒钟内打印该数量时间。

这是主要的

public class main {
    public static void main(String args[]) {
    ScalesSolution s1 = new ScalesSolution(11);
    s1.println();
    ScalesSolution s2 = new ScalesSolution(s1.GetSol());
    s2.println();
}
}

这是 ScalesSolution 类

import java.util.ArrayList;
import java.util.Random;

public class ScalesSolution {
private String scasol;

public void print() {
    System.out.print(scasol);
}

// Display the string with a new line
public void println() {
    print();
    System.out.println();
}



public String GetSol()
{
    return scasol;
}
}

这是 randomOther 类

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

public class randomOther {
// Shared random object
static private Random rand;

// Create a uniformly distributed random integer between aa and bb inclusive
static public int UI(int aa, int bb) {
    int a = Math.min(aa, bb);
    int b = Math.max(aa, bb);
    if (rand == null) {
        rand = new Random();
        rand.setSeed(System.nanoTime());
    }
    int d = b - a + 1;
    int x = rand.nextInt(d) + a;
    return (x);
}

// Create a uniformly distributed random double between a and b inclusive
static public double UR(double a, double b) {
    if (rand == null) {
        rand = new Random();
        rand.setSeed(System.nanoTime());
    }
    return ((b - a) * rand.nextDouble() + a);
}
static public ArrayList<Double> ReadNumberFile(String filename) {
    ArrayList<Double> res = new ArrayList<Double>();
    Reader r;
    try {
        r = new BufferedReader(new FileReader(filename));
        StreamTokenizer stok = new StreamTokenizer(r);
        stok.parseNumbers();
        stok.nextToken();
        while (stok.ttype != StreamTokenizer.TT_EOF) {
            if (stok.ttype == StreamTokenizer.TT_NUMBER) {
                res.add(stok.nval);
            }
            stok.nextToken();
        }
    } catch (Exception E) {
        System.out.println("+++ReadFile: " + E.getMessage());
    }
    return (res);
}
}

这是输出的问题：

00101001010101101011001011010101101001011010001011010010101101001001011010010
01011010010

我相信两个输出应该是相同的，我看到这里有问题，不知道为什么他们不是

4

1 回答 1

0

我看到您在System.out.print内部使用的方式RandomBinaryString(int n)引起了混乱。它正在打印并附加到 String s。尽量避免这种情况。用and替换System.out.print(s += '0');and将修复您的输出。System.out.print(s += '1');s += '0';s += '1';RandomBinaryString

在您的代码中使用以下代码段：

private static String RandomBinaryString(int n) {
    String s = new String();

    // Code goes here
    // Create a random binary string of just ones and zeros of length n
    for (int i = 0; i < n; i++) {
        int y = randomOther.UI(0, 1);
        if (y == 0) {
            s += '0';// this line here was changed
        } else {
            s += '1';// and this line here was changed too
        }
    }

    return (s);
}

希望这可以帮助！

于 2017-02-03T06:02:34.813 回答