如何计算数组中的数字？

Question

我认为我唯一的问题是 displayCounts 方法。我试图让计数数组仅在计数 > 0 时显示计数。计数 > 1 也是如此。所以输入将类似于：2 5 6 5 4 3 4 0。（0 将结束输入）输出将如下所示： - 2 发生 1 次 - 3 次出现 1 次 - 4 出现 2 次 - 5 出现 2 次 - 6 发生 1 次。

public class CountNumbersInArray {
  public static void main(String[] args) {
    java.util.Scanner input = new java.util.Scanner(System.in);
    System.out.print("Enter the integers between 1 and 100: ");

    int[] counts = new int[101];

    int i = 0;
    while (i < counts.length) {
      counts[i] = input.nextInt();
      i++;
    }
    displayCounts(counts);
  }

  public static void displayCounts(int[] counts) {
    int i = 0;
    i++;
    if (counts.length > 0) {
      System.out.println(counts[i] + " occurs " + i + " time");
    } 
    if (counts.length > 1) {
      System.out.println(counts[i] + " occurs " + i + " times");
    }
  }
}

Answer 1

首先，替换

while (i < counts.length) {
  counts[i] = input.nextInt();
  i++;
}

与

i = input.nextInt();
while(i != 0){
  counts[i]++;
  i = input.nextInt();
}

该方法增加用户输入在 counts 数组中的位置的数字，这样数组就会保存数字在特定索引中出现的次数，例如counts[3] 保存 3 出现的频率。
while 循环中的 i != 0 会在输入 0 后立即中断循环并转到下一行的 displayCounts 方法。

基于此，我们现在创建一个新的 displayCounts 方法：

public static void displayCounts(int[] counts){
  for(int i = 0; i < counts.length; i++){
    if(counts[i] > 0){
      System.out.println(i + " occurs " + counts[i] + " times");
    }
  }
}

该方法现在打印一个数字在其计数大于 0 时出现的频率

Answer 2

import java.util.Scanner; // Why not import this class?

public class CountNumbersInArray {
  public static void main(String[] args) {
    Scanner input = new Scanner(System.in).useDelimiter("\\D+");
    System.out.print("Enter the integers between 1 and 100: ");

    int[] counts = new int[101]; // So the last index is 100

    for(int i=0; i < counts.length; i++){
      int number = input.nextInt();
      if(number == 0) break;
      if(number >= 1 && number <= 100) counts[number]++;
      else System.out.println("Your number is outside the boundaries");
    }

    displayCounts(counts);
  }

  public static void displayCounts(int[] counts){
    for(int i = 1; i < counts.length; i++) System.out.println("Number "+i+" occurs "+ counts[i]+ (counts[i] != 1 ? "times" : "time"));
  }
}

这段代码应该可以工作，我所做的是：
1.

for(int i=0; i < counts.length; i++){
  int number = input.nextInt();
  if(number >= 1 && number <= 100) counts[number]++;
  else System.out.println("Your number is outside the boundaries");
}

此代码获取用户输入的数字，并将其计数加一。请注意，这只有效，因为 Integer 的默认值为 0
2.

  public static void displayCounts(int[] counts){
    for(int i = 1; i < counts.length; i++) 
     System.out.println("Number "+i+" occurs "+ counts[i]+ (counts[i] > 1 ? "times" : "time"));
  }

这段代码写得更好。您的代码部分错误是您有 2 个 if 语句，而不是 1 个 if 和 1 个 else if。这样，当计数大于 1 时，您就有 2 个“响应”

Answer 3

试试：

public static void main(String[] args) {
    java.util.Scanner input = new java.util.Scanner(System.in);
    System.out.print("Enter the integers between 1 and 100: ");

    int[] counts = new int[101];
    Arrays.fill(counts, 0);

    int i = 0;
    int inputValue;
    while (i < counts.length) {
        inputValue = input.nextInt();
        counts[inputValue]++;
        i++;
    }
    displayCounts(counts);
}

public static void displayCounts(int[] counts) {
    for (int i = 0; i < counts.length; i++) {
        if (counts[i] > 0) {
            if (counts[i] == 1) {
                System.out.println(i + " occurs " + counts[i] + " time");
            }else {
                System.out.println(i + " occurs " + counts[i] + " times");
            }
        }

    }

}

Answer 4

尝试替换

 while (i < counts.length) {
  counts[i] = input.nextInt();
  i++;
}

与

 int in = input.nextInt();
 while (in != 0) {
   if(in > 0 && in <101){
    counts[in]++;
   } else {
     System.out.println("Please enter values between 1-100. Enter 0 to 
  stop");
  }
 in = input.nextInt();
}

然后遍历整个计数数组以获得 0 到 100 之间元素的各个频率