Question

此Java程序中没有任何问题，唯一的问题是我想知道是否会有更简单的方法仅使用for循环来完成此操作，因为我刚刚开始学习for Java中的-loops。以一种更简单的方式，我的意思是一种使用更少的代码的更有效的方法。

import java.util.Scanner;
public class BarChart 
{
    public static void main(String[] args)
    {
        //#Declaring variables
        int sales1, sales2, sales3, sales4, sales5;

        //#Creating scanner object
        Scanner keyboard = new Scanner(System.in);  

        //#Asking user for input
        System.out.print("Enter today's sales for store 1: ");
        sales1 = keyboard.nextInt();

        System.out.print("Enter today's sales for store 2: ");
        sales2 = keyboard.nextInt();

        System.out.print("Enter today's sales for store 3: ");
        sales3 = keyboard.nextInt();

        System.out.print("Enter today's sales for store 4: ");
        sales4 = keyboard.nextInt();

        System.out.print("Enter today's sales for store 5: ");
        sales5 = keyboard.nextInt();

        //#Displaying the sales bar chart
        System.out.println("\nSALES BAR CHART");

        System.out.print("\nStore 1: ");        
        for (int num = 0; num < sales1; num += 100)
        {
            System.out.print("*");
        }

        System.out.print("\nStore 2: ");
        for (int num = 0; num < sales2; num += 100)
        {
            System.out.print("*");
        }

        System.out.print("\nStore 3: ");
        for (int num = 0; num < sales3; num += 100)
        {
            System.out.print("*");
        }

        System.out.print("\nStore 4: ");
        for (int num = 0; num < sales4; num += 100)
        {
            System.out.print("*");
        }

        System.out.print("\nStore 5: ");
        for (int num = 0; num < sales5; num += 100)
        {
            System.out.print("*");
        }
    }
}

输出应如下所示：

> What the Output should look like: 
> Enter today's sales for store 1: 1000 
> Enter today's sales for store 2: 1290 
> Enter today's sales for store 3: 1850 
> Enter today's sales for store 4: 800 
> Enter today's sales for store 5: 1900

> SALES BAR CHART 
> Store 1: ********** 
> Store 2: ************* 
> Store 3: ******************* 
> Store 4: ******** 
> Store 5: *******************

Answer 1

当然，您可以遍历大多数重复性任务：

public static void main(String[] args){
    int[] sales = new int[5];
    Scanner keyboard = new Scanner(System.in); 
    //Ask the user for input five times, stored the input in the sales array
    for (int i = 0; i < 5; i++){
        System.out.print("Enter today's sales for store " + (i+1) ": ");
        sales[i] = keyboard.nextInt();
    }
    //Print the sales
    System.out.println("Sales bar chart");
    for (int i = 0; i < 5; i++){
        printStore(i, sales[i]);
    }
}

//helper function that prints a resulting store line
private void printStore(int storeNumber, int sales){
    System.out.print("Store " + (i + 1) + ": "):
    for (int i = 0; i < sales / 100; i++){
        system.out.print("*");
    }
    System.out.println();
}

请注意，尽管这减少了代码行，但并没有比您的解决方案更有效。

Answer 2

每次编写同一行代码超过三遍时，都应考虑使用循环。同样，只要发现使用foo1，foo2等变量名，您就应该考虑将它们放入数据结构中。如果您使用简单的Array并进行for循环，则程序会简化为：

int[] stores = new int[5];
Scanner in = new Scanner(System.in);
for(int i = 0; i < stores.length; i++) {
    System.out.print("Enter today's sales for store " + (1+i) + ": ");
    stores[i] = in.nextInt();
}
for(int i = 0; i < stores.length; i++) {
    System.out.print("Store " + (i+1) + ": ");
    for(int j =0; j < stores[i]; j++) {
        System.out.print("*");
    }
    System.out.println();
}

样品运行：

Enter today's sales for store 1: 3
Enter today's sales for store 2: 5
Enter today's sales for store 3: 7
Enter today's sales for store 4: 2
Enter today's sales for store 5: 8
Store 1: ***
Store 2: *****
Store 3: *******
Store 4: **
Store 5: ********

Answer 3

使用基本编码进行编码的一种简单方法是仅在条件条件下使用，并每次将*附加到String上，因此您只需要使用一个循环。这是下面的样子。我没有改变您从用户那里获取值的方式。

String count = "", count2 = "", count3 = "", count4 = "", count5 = ""; int max = Collections.max(Arrays.asList(sales1,sales2,sales3,sales4,sales5)); for (int num = 0; num < max; num += 100) { if (num < sales1) { count = count + "*"; } if (num < sales2) { count2 = count2 + "*"; } if (num < sales3) { count3 = count3 + "*"; } if (num < sales4) { count4 = count4 + "*"; } if (num < sales5) { count5 = count5 + "*"; } } System.out.print("\nStore 1: " + count); System.out.print("\nStore 2: " + count2); System.out.print("\nStore 3: " + count3); System.out.print("\nStore 4: " + count4); System.out.print("\nStore 5: " + count5);

int max用于查找5个销售数字中的最大值，以便循环正确退出。由于在循环的同一遍中覆盖了多个*，因此在速度上也是一种更有效的方法。我故意没有不调用方法或使用更复杂的方法来存储值，以使您始终专注于循环，因为您当前正在学习循环。

Answer 4

如果您想尝试使用lambda（Java 8）

    public static void main(String[] args) {
        // Declaring steps
        int N = 5;
        // Declaring increment
        int INCREMENT = 100;
        // Creating scanner object
        Scanner keyboard = new Scanner(System.in);

        // Creating a loop from 1 to N (including the limits)
        List<Integer> variables = IntStream.rangeClosed(1, N)
                // Asking user for input
                .peek(index -> System.out.printf("Enter today's sales for store %d: ", index))
                // Read the user answer
                .mapToObj(dummy -> keyboard.nextInt())
                // Collect all the answers in a list
                .collect(Collectors.toList());

        // Displaying the sales bar chart
        System.out.println("\nSALES BAR CHART\n");

        // Creating a loop from 0 to N (without N)
        IntStream.range(0, N)
                // Display the message for each step
                .peek(index -> System.out.printf("Store %d: ", index + 1))
                // Get the saved value
                .map(variables::get)
                // Display the stars
                .forEach(element -> System.out.println("*".repeat(element / INCREMENT)));
    }

对于此星号条形图Java程序，还有其他更简单的方法吗？

4 个答案: