从循环中打印

Printing from a loop

我需要打印一个完美数的因数。这是我的主要内容 class:

ArrayList<Integer> perfNums = new ArrayList<>();
Scanner in = new Scanner(System.in);
System.out.print("Enter the upperbound: ");
upperbound = in.nextInt();
for (int i = 1; i <= upperbound; i++) {
     if (isPerfect(i)) { //boolean to check if number is a perfect number
        perfNums.add(i);
     }
}

System.out.println("Perfect numbers between 1 and " + upperbound + " are:");
for (int i = 0; i < perfNums.size(); i++) {
     System.out.print(perfNums.get(i) + " = ");
     printFactor((int)perfNums.get(i));
     System.out.println();
}

这是 printFactor class.

private static void printFactor(int number){
    int factor = 1;
    while(factor < number){
        if (number%factor == 0) System.out.print(factor+ " + ");
        //I don't know how to print the + sign otherwise.
        factor++;
    }
}

这是一个示例输出:

Enter the upperbound: 10000
Perfect numbers between 1 and 10000 are:
6 = 1 + 2 + 3 + 
28 = 1 + 2 + 4 + 7 + 14 + 
496 = 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 + 
8128 = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 127 + 254 + 508 + 1016 + 2032 + 4064 + 

我已经掌握了它的要点,但我一直在努力解决输出问题。由于我的在线提交系统的限制,我的输出需要符合确切的规格。

我的问题是如何打印完全数的所有因式但去掉末尾的 + 号?(例如)6 = 1 + 2 + 3

我不太确定从 while 循环打印的许多方法。 for 循环是否更适合我的目标?或者是否有其他方法来打印数字的因数?

您应该将要打印的输出缓存到 StringBuilder 中。然后您可以在打印整个 String 之前删除最后一个加号。它还具有更好的性能。

private static void printFactor(int number)
{
    StringBuilder output = new StringBuilder();
    int factor = 1;
    while (factor < number)
    {
        if (number % factor == 0)
            output.append(factor + " + ");
        factor++;
    }
    // remove last plus sign
    output.deleteCharAt(output.length() - 1);
    // print the whole string
    System.out.print(output.toString());
}

解决这个问题的最少更改可能是这样的:

private static void printFactor(int number) 
    System.out.print(1);
    int factor = 2;
    while (factor<number) {
        if (number%factor == 0) System.out.print(" + " + factor);
        factor++;
    }
}

1 始终是一个因素,因此您可以在循环之前打印它,然后将 + 添加到每个后续因素。

由于 factor 从值 1 开始并且 number % 1 == 0 将始终为真,您可以先打印 1 然后翻转 factor+ in System.out.print.像这样:

private static void printFactor(int number) {
    if(number > 0) {
        System.out.print(1);
    }
    int factor = 2;
    while (factor<number) {
        if (number % factor == 0) {
            System.out.print(" + " + factor);
        }
        factor++;
    }
}

不是最好的解决方案,但它可以完成工作。

尝试创建一个变量字符串 numb 并使用 substring 方法,如下所示:

      String numb ="";
        while(factor<number){
            if(number%factor == 0) 
                numb= numb + factor+ " + ";
            factor++;
        }
        System.out.print(numb.substring(0, numb.trim().length()-1));

感谢大家的快速回复。你们都是救命恩人,我在以后编写代码时设法挑选了一些新的东西来考虑。

无论如何,在等待回复的过程中,我一直在摆弄代码,并想出了一个相当不优雅的解决方案,如果有人感兴趣的话。这是对主要 class:

的更改
System.out.println("Perfect numbers between 1 and " + upperbound + " are:");
     for(int i=0; i<perfNums.size(); i++){
         System.out.print(perfNums.get(i) + " = ");
         outputString = printFactor2(perfNums.get(i));
         if(outStr.endsWith(" + ")) outStr = outStr.substring(0, outStr.length()-3); 
         //because the submission system would cry foul with even a single extra space
         System.out.println(outStr);
     }

下面是对 printFactor class 的更改:

private static String printFactor2(int number){
    String out = "";
    int factor = 1;
    while(factor<number){
        if(number%factor == 0) out += factor + " + ";
        factor++;
    }
    return out;
}

基本上,我所做的是将因子附加到字符串,然后使用子字符串方法删除尾随的 + 号。事后看来,我可能应该改为在 printFactor class 中调用 substring 方法。也许像 return out.substring(0, out.length()-3);

不过还是谢谢大家!

只是为了使用 Java 8 :)

private static void printFactor(int number){
    System.out.println(IntStream.range(1, number)
                                .filter(p -> number % p == 0)
                                .mapToObj(i -> String.valueOf(i))
                                .collect(Collectors.joining(" + ")));
}