Question

我的Java程序存在问题。我创建了一个程序，它接受一个整数并将其转换为二进制值。当数字是奇数时，没有任何问题。 15转换为1111,17转换为10001，依此类推。当数字是偶数时，就会出现问题。如果我输入16,18,20等，它只返回0.每次。另外，重要的是要注意我通过使用递归方法得到我的号码，一旦它到达它的号码就会停止。

这是我的代码。感谢我能得到的任何帮助，即使它没有解决问题。

public class binaryConverter {
    static int nr = 16;
    static int max = 0;
    static int[] bin;
    static int[] array;
    static int finalBin;

    public static void main(String[] args) {
        maxFinder();    
        binMaker();     
        toBinary(array, 0,true);
    }

    //finds out how many binary numbers are used in order to decide what length to make the array, 
    //15 = 1111, 10000 = 16 15<16
    private static void maxFinder(){    
        for(int i = 0, n = 1; i<nr; i++){
            if(n>nr){
                max = i;
                break;
            }
            n*=2;       //n doubles for every i loop, starts with one
        }
    }

    //makes a library of standard to binary (0 = 1, 1 = 2; 2 = 4; 3 = 8...)
    private static void binMaker(){     
        int[] temp = new int[max];
        for(int i = 0; i<temp.length; i++){
            if(i == 0) temp[i] = 1;
            else temp[i]=2*temp[i-1];
        }
        bin = temp;
        array = new int[bin.length];
    }

    //adds the array together in order to access what number the array currently resembles in binary
    private static int sum(int[] ar, int length){
        int sum = 0;
        for(int i = 0; i<=length; i++) if(ar[i]==1) sum += bin[i];
        return sum;
    }

    //loops until the array becomes the number in binary
    private static void toBinary(int[] ar, int i, boolean one){     //i = the current number it's on, eg. 10i01, i is the third slot
        if(i==array.length) return;     //break if 
        ar[i] = (one) ? 1:0;
        if(sum(ar, i)==nr){     //if the temporary array is the number but in binary ...
            array = ar;     //turns the static array into the temporary array
            String temp = "";
            for(int z = 0; z<array.length; z++) temp += array[z];
            finalBin = Integer.parseInt(temp);  //makes finalBin represent the original number but in binary
            return;
        }
        else{   //else go to the next slot
            toBinary(ar, i+1, true);
            toBinary(ar, i+1, false);   
        }
    }
}

编辑：我现在已将以下行添加到我的主页： if（finalBin！= nr）toBinary（array，0，false）; 的System.out.println（finalBin）; 这是为了确保它可以从0开始。但是，我仍然得到不正确的答案，因为它给我看起来偶数回报偶数。

Answer 1

你开始递归时总是在二进制文件的第一位有一个：

toBinary(array, 0, true);

这样你永远不会得到偶数。偶数在“第一”位始终为零（表示“2为0的幂”）。

您可以像这样开始递归：

toBinary(array, 0, true);

if (/* not found a solution */)
   toBinary(array, 0, false);

Answer 2

您可以将此代码用作转换器，并将String类型替换为某些List：

public static String decToBin(int value) {
    String result = "";
    while (value > 1) {
        result += value % 2;
        value /= 2;
    }
    result += value;
    result = new StringBuilder(result)
            .reverse()
            .toString();
    return result;
}

Answer 3

这就是你如何让它发挥作用：

public class binaryConverter {
    static int nr = 16;
    static int max = 0;
    static int[] bin;
    static int[] array;
    static int finalBin;
    static boolean foundSolution = false;

    public static void main(String[] args) {
        maxFinder();
        binMaker();
        toBinary(array, 0, true);

        if (!foundSolution)
            toBinary(array, 0, false);

        for (int i = array.length - 1; i >= 0; i--)
            System.out.print(array[i]);
        System.out.println();
    }

    //finds out how many binary numbers are used in order to decide what length to make the array,
    //15 = 1111, 10000 = 16 15<16
    private static void maxFinder(){
        for(int i = 0, n = 1; i<nr; i++){
            if(n>nr){
                max = i;
                break;
            }
            n*=2;       //n doubles for every i loop, starts with one
        }
    }

    //makes a library of standard to binary (0 = 1, 1 = 2; 2 = 4; 3 = 8...)
    private static void binMaker(){
        int[] temp = new int[max];
        for(int i = 0; i<temp.length; i++){
            if(i == 0) temp[i] = 1;
            else temp[i]=2*temp[i-1];
        }
        bin = temp;
        array = new int[bin.length];
    }

    //adds the array together in order to access what number the array currently resembles in binary
    private static int sum(int[] ar, int length){
        int sum = 0;
        for(int i = 0; i<=length; i++) if(ar[i]==1) sum += bin[i];
        return sum;
    }

    //loops until the array becomes the number in binary
    private static void toBinary(int[] ar, int i, boolean one){     //i = the current number it's on, eg. 10i01, i is the third slot
        if(i==array.length || foundSolution) return;     //break if
        ar[i] = (one) ? 1:0;
        if(sum(ar, i)==nr){     //if the temporary array is the number but in binary ...
            array = ar;     //turns the static array into the temporary array
            String temp = "";
            for(int z = 0; z<array.length; z++) temp += array[z];
            finalBin = Integer.parseInt(temp);  //makes finalBin represent the original number but in binary
            foundSolution = true;
            return;
        }
        else{   //else go to the next slot
            toBinary(ar, i+1, true);
            toBinary(ar, i+1, false);
        }
    }
}

在奇数上成功转换为二进制，在偶数上失败

3 个答案: