Question

我需要将min和max值存储在给定的数组中，然后使用特定字符打印出来（+表示最小值，“ - ”表示最小值，“*”表示所有其余值）。 / p>

我认为除了适当的存储值之外，我已经完成了大部分操作，因此所有值都不是“++++++++++”....就像它们当前打印出来一样。

有什么想法吗？非常感谢帮助。

      public static void main(String[] args) {
        Scanner keyboard = new Scanner(System.in);
        int numbers[] = new int[24];
        int min = Integer.MAX_VALUE;
        int max=Integer.MIN_VALUE;
        int maxhour = 0; 
        int minhour = 0;
        int total = 0;
        char MAX = '+', MIN = '-', MIDDLE = '*';
        char currentchar = 0;
        for(int i=0; i< numbers.length; i++){
            numbers[i] = keyboard.nextInt();
            total = total + numbers[i];
           if(numbers[i]<min){
               min = numbers[i];
               minhour = i;
               currentchar = MIN;
           }else if (numbers[i]>max){
               max = numbers[i];
               maxhour = i;
               currentchar = MAX;
           }
        }
        for(int i=0; i< numbers.length; i++){
            System.out.print("Hour " + i + ":");
            printTimes(currentchar, numbers[i]);
            System.out.println("");
          }

        System.out.println("Largest Number of hits is : " + max + " at hour " + maxhour);
        System.out.println("Average Number of hits is : " + (total/24) + " per hour");
        System.out.println("Smallest Number of hits is : " + min + " at hour " + minhour);



    }

        public static void printTimes(char c, int times) {
        if (times >= 70) {
            for(int i=0; i< 69; i++){
            System.out.print(c);
            } System.out.print(">");
        } else if (times < 70) {
            for(int i=0; i< times; i++)
            System.out.print(c);
        }
    }


}

当前输出示例：

42 29 36 7 5 3 10 13 33 40 51 49 22 58 63 102 65 58 48 24 36 48 52 42

小时0：++++++++++++++++++++++++++++++++++++++++++++++++++++++ p>

第1小时：++++++++++++++++++++++++++++++++++++

第2小时：+++++++++++++++++++++++++++++++++++++++++

第3小时：+++++++

第4小时：+++++

第5小时：+++

第6小时：+++++++++++++++++++++

第7小时：++++++++++++++++++++

第8小时：+++++++++++++++++++++++++++++++++++++++

第9小时：++++++++++++++++++++++++++++++++++++++++++++++

...

最大点击次数是：小时15点

平均点击次数为：每小时39次

最小点击次数为：3时5点

Answer 1

只需更改上一个for：

for (int i = 0; i < numbers.length; i++) {
    System.out.print("Hour " + i + ":");
    if (numbers[i] == min)
        currentchar = MIN;
    else if (numbers[i] == max)
        currentchar = MAX;
    else
        currentchar = MIDDLE;
    printTimes(currentchar, numbers[i]);
    System.out.println("");
}

Answer 2

我会按如下方式更新你的代码 -

public static void main(String[] args) {
  int numbers[] = new int[24];
  int total = 0;

  System.out.println("Enter 24 integers please");
  Scanner keyboard = null;
  try {
    keyboard = new Scanner(System.in);
    for (int i = 0; i < numbers.length; i++) {
      numbers[i] = keyboard.nextInt();
      total += numbers[i];
    }
  } finally {
    keyboard.close();
  }
  Integer min = null;
  Integer max = null;
  int maxhour = 0;
  int minhour = 0;
  char MAX = '+', MIN = '-', MIDDLE = '*';
  for (int i = 0; i < numbers.length; i++) {
    if (min == null || numbers[i] < min) {
      min = numbers[i];
      minhour = i;
    } else if (max == null || numbers[i] > max) {
      max = numbers[i];
      maxhour = i;
    }
  }
  for (int i = 0; i < numbers.length; i++) {
    char currentchar = MIDDLE;
    if (i == minhour) {
      currentchar = MIN;
    } else if (i == maxhour) {
      currentchar = MAX;
    }
    System.out.print("Hour " + i + ":");
    printTimes(currentchar, numbers[i]);
    System.out.println("");
  }

  System.out.println("Largest Number of hits is : "
    + max + " at hour " + maxhour);
  System.out.println("Average Number of hits is : "
    + (total / 24) + " per hour");
  System.out.println("Smallest Number of hits is : "
    + min + " at hour " + minhour);
}

JAVA - 如何在循环内的数组中存储最大值和最小值？

2 个答案: