Java：将整数数字从大到小排序

Question

我需要帮助仅使用 if 语句而不使用 String/Arrays 或任何类似的东西来对整数数字从大到小进行排序。假设我已经初始化并导入了 Scanner 和键盘以允许输入。我遇到的问题是 if 语句，因为我似乎无法正确处理。请帮忙，过去 5 个小时我一直在努力解决这个问题。谢谢:)

    System.out.print("Enter a five digit integer number: ");

    int fiveInt, digit1, digit2, digit3, digit4, digit5; //Declaring variables

    Scanner keyboard = new Scanner (System.in);

    fiveInt = keyboard.nextInt(); //User input will be required



    System.out.println(" "); //Prints empty line


    //The following will be explained assuming number entered was 12345.
    //The (int) will change whatever result of the division taking place to an integer.

    digit1 = (int)(fiveInt / 10000); //12345 divided by 10000 = 1.2345 converted to 1 due to (int).
    digit2 = ((int)(fiveInt / 1000)) - (digit1 * 10); //12345 divided by 1000 = 12.345 - (1 * 10) = 2.345 converted 2 due to (int).
    digit3 = ((int)(fiveInt / 100)) - (digit1 * 100) - (digit2 * 10); //12345 divided by 100 = 123.45 - (1*100) - (2*10) = 3.45 converted to 3 due to (int).
    digit4 = ((int)(fiveInt / 10)) - (digit1 * 1000) - (digit2 * 100) - (digit3 * 10); //12345 divided by 10 = 1234.5 - (1*1000) - (2*100) - (3*10) = 4.5 converted to 4 due to (int).
    digit5 = fiveInt - (digit1 * 10000) - (digit2 * 1000) - (digit3 * 100) - (digit4 * 10); //12345 - (1*10000) - (2*1000) - (3*100) - (4*10) = 5



    System.out.println("The digits in " + fiveInt + " are: " + digit1 + ", " + digit2 + ", " + digit3 + ", " + digit4 + ", " + digit5);

    System.out.println(" "); //Prints empty line




    //Insert explanation for lines of code below here.



    if(digit1 < digit2){
      int a = digit1;
      digit1 = digit2;
      digit2 = digit1;
    }

    if(digit1 < digit3){
      int b = digit1;
      digit1 = digit3;
      digit3 = digit1;
    }

    if(digit1 < digit4){
      int c = digit1;
      digit1 = digit4;
      digit4 = digit1;
    }

    if(digit1 < digit5){
      int d = digit1;
      digit1 = digit5;
      digit5 = digit1;
    }




    //Insert explanation for lines of code above here.





    System.out.print("The largest number with these digits is: " + digit1 + digit2 + digit3 + digit4 + digit5); //Displays the digits in their sorted mann

Answer 1

第一个错误是您没有恢复临时存储在 a、b、c...而不是

中的变量

if(digit1 < digit2){
      int a = digit1;
      digit1 = digit2;
      digit2 = digit1;
    }

你可能想要：

if(digit1 < digit2){
      int a = digit1;
      digit1 = digit2;
      digit2 = a;
    }

无论如何，这还不够，因为算法不会 'sort' 这五个变量。使用此代码，您只需将最大数字存储在 digit1.

中

如果你想要真正的排序，最简单的方法是将整数放入数组中并使用 Arrays.sort()

对其进行排序

另外，你的最后一行代码：

System.out.print("The largest number with these digits is: " + digit1 + digit2 + digit3 + digit4 + digit5);

不会打印排序的整数，而是数字的总和（在您的示例中为 15）

Answer 2

您需要定义 5 项，例如 a1、a2、a3、a4、a5，然后找到最大或最小的排序项

Answer 3

这是我想出的。基本思想是找到最小的数字，然后使这个数字无效。不幸的是，这意味着原始数字数据丢失了，但您始终可以从用户输入中恢复它 int:

int currentMin;

System.out.print("The largest number with these digits is: ");

while(true) {
  currentMin = Math.min(digit1, Math.min(digit2, Math.min(digit3, Math.min(digit4, digit5))));
  //Invalidate the min digit
  if (currentMin == digit1) digit1 = Integer.MAX_INT;
  if (currentMin == digit2) digit2 = Integer.MAX_INT;
  if (currentMin == digit3) digit3 = Integer.MAX_INT;
  if (currentMin == digit4) digit4 = Integer.MAX_INT;
  if (currentMin == digit5) digit5 = Integer.MAX_INT;
  if (currentMin == Integer.MAX_INT) {
    break;
  } else {
    System.out.print(currentMin + " ");
  }
}

Answer 4

这是对我刚刚写的内容的阐述。基本上你必须得到 5 个整数（a1 到 a5）和 returns 排序的值。我先找到顺序。例如在下面的测试数据中，7 是第 3 项。然后将 7 和其他人按照我提到的从 a1 到 a5 的预定义整数的适当顺序打印出来 ...

public class B{

static int a1, a2, a3, a4, a5;

static int digit1=4;
static int digit2=1;
static int digit3=7;
static int digit4=9;
static int digit5=5;

public static void main(String args[]) {


    setDigitOrer(digit1, getOrder(digit1, digit2, digit3, digit4, digit5));
    setDigitOrer(digit2, getOrder(digit2, digit1, digit3, digit4, digit5));
    setDigitOrer(digit3, getOrder(digit3, digit2, digit1, digit4, digit5));
    setDigitOrer(digit4, getOrder(digit4, digit2, digit3, digit1, digit5));
    setDigitOrer(digit5, getOrder(digit5, digit2, digit3, digit4, digit1));


    System.out.println(a1 );
    System.out.println(a2 );
    System.out.println(a3 );
    System.out.println(a4 );
    System.out.println(a5 );

}

private static void setDigitOrer(int digit, int digitOrder) {
    if (digitOrder == 0){
        a1 = digit;
    } else if (digitOrder == 1){
        a2 = digit;
    } else if (digitOrder == 2){
        a3 = digit;
    } else if (digitOrder == 3){
        a4 = digit;
    } else if (digitOrder == 4){
        a5 = digit;
    }

}

private static int getOrder(int digit, int... digits){
    int count = 0;
    for (int d: digits){
        if (d > digit){
            count++;
        }
    }
    return count;
}

}

结果是 9 7 5个 4个 1

Answer 5

这绝对是作业，规则不是很清楚。但是，我确信该解决方案不包括对 Math.min 或 Math.max 方法的 4 次调用。动动脑筋，查找一些排序算法并尝试实现一个。

Answer 6

使用以下 python 小程序：

for pos in range(1,5):
    for num in range(1,6-pos):
        print "if (digit%d < digit%d) {" % (num, num+1)
        print "\tSystem.out.println(\"Swapping \"+digit%d+\" and \"+digit%d);" % (num, num+1)
        print "\tint temp=digit%d;" % (num)
        print "\tdigit%d=digit%d;" % (num, num+1)
        print "\tdigit%d=temp;" % (num+1)
        print "}"

它将生成 java 代码，您可以在这些注释之间插入代码

//Insert explanation for lines of code below here.

它将对您的五位数字进行排序。基本上它 bubble sort 展开两个循环。

不要浪费时间手动输入排序语句。确实存在计算机来为我们完成无聊的工作！

if (digit1 < digit2) {
    System.out.println("Swapping "+digit1+" and "+digit2);
    int temp=digit1;
    digit1=digit2;
    digit2=temp;
}
if (digit2 < digit3) {
    System.out.println("Swapping "+digit2+" and "+digit3);
    int temp=digit2;
    digit2=digit3;
    digit3=temp;
}
...

Answer 7

    System.out.print("Enter a five digit integer number: ");

    int fiveInt, digit1, digit2, digit3, digit4, digit5; // Declaring
                                                            // variables

    Scanner keyboard = new Scanner(System.in);

    fiveInt = keyboard.nextInt(); // User input will be required

    System.out.println(" "); // Prints empty line

    // The following will be explained assuming number entered was 12345.
    // The (int) will change whatever result of the division taking place to
    // an integer.

    digit1 = (int) (fiveInt / 10000); // 12345 divided by 10000 = 1.2345
                                        // converted to 1 due to (int).
    digit2 = ((int) (fiveInt / 1000)) - (digit1 * 10); // 12345 divided by
                                                        // 1000 = 12.345 -
                                                        // (1 * 10) = 2.345
                                                        // converted 2 due
                                                        // to (int).
    digit3 = ((int) (fiveInt / 100)) - (digit1 * 100) - (digit2 * 10); // 12345
                                                                        // divided
                                                                        // by
                                                                        // 100
                                                                        // =
                                                                        // 123.45
                                                                        // -
                                                                        // (1*100)
                                                                        // -
                                                                        // (2*10)
                                                                        // =
                                                                        // 3.45
                                                                        // converted
                                                                        // to
                                                                        // 3
                                                                        // due
                                                                        // to
                                                                        // (int).
    digit4 = ((int) (fiveInt / 10)) - (digit1 * 1000) - (digit2 * 100)
            - (digit3 * 10); // 12345 divided by 10 = 1234.5 - (1*1000) -
                                // (2*100) - (3*10) = 4.5 converted to 4 due
                                // to (int).
    digit5 = fiveInt - (digit1 * 10000) - (digit2 * 1000) - (digit3 * 100)
            - (digit4 * 10); // 12345 - (1*10000) - (2*1000) - (3*100) -
                                // (4*10) = 5

    System.out
            .println("The digits in " + fiveInt + " are: " + digit1 + ", "
                    + digit2 + ", " + digit3 + ", " + digit4 + ", "
                    + digit5);

    System.out.println(" "); // Prints empty line

    // Insert explanation for lines of code below here.

    if (digit1 < digit2) {
        int a = digit1;
        digit1 = digit2;
        digit2 = a;
    }

    if (digit2 < digit3) {
        int b = digit2;
        digit2 = digit3;
        digit3 = b;
    }

    if (digit3 < digit4) {
        int c = digit3;
        digit3 = digit4;
        digit4 = c;
    }

    if (digit4 < digit5) {
        int d = digit4;
        digit4 = digit5;
        digit5 = d;
    }

    // 2nd Filtering

    if (digit1 < digit2) {
        int a = digit1;
        digit1 = digit2;
        digit2 = a;
    }

    if (digit2 < digit3) {
        int b = digit2;
        digit2 = digit3;
        digit3 = b;
    }

    if (digit3 < digit4) {
        int c = digit3;
        digit3 = digit4;
        digit4 = c;
    }
    // 3rd Filtering

    if (digit1 < digit2) {
        int a = digit1;
        digit1 = digit2;
        digit2 = a;
    }

    if (digit2 < digit3) {
        int b = digit2;
        digit2 = digit3;
        digit3 = b;
    }

    // last filtering
    if (digit1 < digit2) {
        int a = digit1;
        digit1 = digit2;
        digit2 = a;
    }

    // Insert explanation for lines of code above here.

    System.out.print("The largest number with these digits is: " + digit1
            + digit2 + digit3 + digit4 + digit5); // Displays the digits in
                                                    // their sorted mann

Answer 8

我在 Java 中编写了一个算法，它在 冒泡排序 模型算法和 中解决了这个问题，而不用分隔数组或变量中的数字.

在我看来，这是最干净的解决方案，可能是最快的解决方案（如果有人能真正对这种算法进行基准测试，我将不胜感激）。

我最初创建这个算法是因为我很惊讶没有人试图解决这个问题这个方法而是将每个数字存储在单独的变量中，这会占用更多内存，在我看来是一种糟糕的编程和思维方式，因为它包含固定长度的参数整数等边界。

Fork it @ Github

public static int main(int n) {
    int l, r;
    final int size = (int)(Math.log10(n));
    for (int i = 0; i < size; i++) 
        for (int k = i; k < size; k++) {
            r = (int)(n/Math.pow(10,k)%10);
            l = (int)(n/Math.pow(10,k+1)%10);
            if(l < r)
                n += Math.pow(10,k)*(l-r) + Math.pow(10,k+1)*(r-l);
        }
    return n;
}

Java：将整数数字从大到小排序

Java: Sorting integer digits from greatest to least

java

sorting

if-statement