在 Java 中，为什么我不能从超过 255 的十进制数中获取 9 个字符？是符号位吗？

Question

这是针对 headtails java 作业分配的，您可以在网上找到一些地方（例如http://www.javaproblems.com/2013/01/medium-problem-tricky-heads-and-tails.html）这个想法是输入一个十进制 0 到 511 并让它输出一个 3 x 3 的 H 或 T 矩阵，用于 0 或 1（我的作品是 0 到 255）这是我的尝试，但我无法开始工作：public static void main(String[] args) {    @SuppressWarnings("resource")    Scanner keyboard = new Scanner(System.in);    //System.out.println("Please enter a number between 0 and 511: ");    //int num = keyboard.nextInt();    int num = 458;    String binNum = "";    int temp;    String[][] coinArr = new String[3][3]; // = [][];    while(num > 0)    {        temp = (int) (num % 2);        binNum = binNum + "" + temp;        num = (int) (num / 2);    }    System.out.println("binNum length is " + binNum.length());    System.out.println("binNum is " + binNum);    binNum = String.format((binNum.length() < 9 ? ("%0"+ (9 - binNum.length())+"d%s") : "%0$d%s"), 0 ,binNum);    System.out.println("binNum length is " + binNum.length());    System.out.println("binNum is " + binNum);    int k=0;    for(int i = 0; i < 3; i++) {        for (int j = 0; j < 3; j++) {            binNum = binNum.replaceAll("0", "H");            binNum = binNum.replaceAll("1", "T");            coinArr[i][j] = binNum.substring(k, k+1);            k++;            System.out.print(coinArr[i][j]);        }        System.out.println();    }}

Answer 1

您从十进制到二进制的转换是错误的，正确的是：

while(num > 0)

{

    temp = (int) (num % 2);

    binNum = temp + "" + binNum;

    num = (int) (num / 2);

}

那么你的 binNum 只需要 9 个字符：

binNum = binNum.length() < 9 ? String.format("%0"+ (9 - binNum.length())+"d%s", 0 ,binNum) : binNum;

// You don't need this inside your loop

binNum = binNum.replaceAll("0", "H");

binNum = binNum.replaceAll("1", "T");

int k=0;

for(int i = 0; i < 3; i++) {

    for (int j = 0; j < 3; j++) {

        coinArr[i][j] = String.valueOf(binNum.charAt(k));

        k++;

        System.out.print(coinArr[i][j]);

    }

    System.out.println();

}

Answer 2

有2个问题：-

1) 改变

binNum = binNum + "" + temp;

到

binNum = temp + binNum;

2) 改变

binNum = String.format((binNum.length() < 9 ? ("%0"+ (9 - binNum.length())+"d%s") : "%0$d%s"), 0 ,binNum);

到

binNum = String.format("%09d", Integer.parseInt(binNum));