Question

我的程序从文件中读取任意数量的整数，然后打印1到100之间所有整数的直方图条形图。代码工作，我已经尝试使其缩短所有可能的格式，但然后我的代码停止工作。所以这个长版本是目前唯一可用的版本。所以我的问题只是出于好奇，如果我的if语句和直方图的打印有一个较短的方式。
注意：并非文件中的所有整数都必须位于区间[1-100]

中

    Create an object representing a file
    File file = new File("path");
    Scanner fileScan = new Scanner(file);

    ArrayList<Integer> list = new ArrayList<Integer>();
    int total=0;
    while (fileScan.hasNext()){
        total++;
            list.add(fileScan.nextInt());
    }
    int [] counter = new int [10];
    for(int i=0; i<list.size();i++){
        if (list.get(i) >=1 && list.get(i)<=10){
            counter[0]++;
        }
        if (list.get(i) >10 && list.get(i)<=20){
            counter[1]++;
        }
        if (list.get(i) >20 && list.get(i)<=30){
            counter[2]++;
        }
        if (list.get(i) >30 && list.get(i)<=40){
            counter[3]++;
        }
        if (list.get(i) >40 && list.get(i)<=50){
            counter[4]++;
        }
        if (list.get(i) >50 && list.get(i)<=60){
            counter[5]++;
        }
        if (list.get(i) >60 && list.get(i)<=70){
            counter[6]++;
        }
        if (list.get(i) >70 && list.get(i)<=80){
            counter[7]++;
        }
        if (list.get(i) >80 && list.get(i)<=90){
            counter[8]++;
        }
        if (list.get(i) >90 && list.get(i)<=100){
            counter[9]++;
        }
    }
    int valueTotal=0;
    for (int j=0; j<counter.length; j++){
        valueTotal += counter[j];
    }
    System.out.print("Reading integers from file: "+file);
    System.out.print("\nNumber of integers in the interval [1,100]: "+valueTotal);
    System.out.print("\nOthers: "+(total-valueTotal));
    System.out.print("\nHistogram\n");  
    System.out.print("1 - 10  | ");
    display(counter[0]);
    System.out.print("\n11 - 20 | ");
    display(counter[1]);
    System.out.print("\n21 - 30 | ");
    display(counter[2]);
    System.out.print("\n31 - 40 | ");
    display(counter[3]);
    System.out.print("\n41 - 50 | ");
    display(counter[4]);
    System.out.print("\n51 - 60 | ");
    display(counter[5]);
    System.out.print("\n61 - 70 | ");
    display(counter[6]);
    System.out.print("\n71 - 80 | ");
    display(counter[7]);
    System.out.print("\n81 - 90 | ");
    display(counter[8]);
    System.out.print("\n91 - 100| ");
    display(counter[9]);
    }
    public static void display(int n){
    for (int i=0; i<n; i++){
        System.out.print("*");
    }
    }
}

我的输出：

   Reading integers from file: ....txt
   Number of integers in the interval [1,100]: 18
   Others: 4
   Histogram
   1 - 10  | ******
   11 - 20 | *
   21 - 30 | ***
   31 - 40 | 
   41 - 50 | *
   51 - 60 | *
   61 - 70 | *
   71 - 80 | ***
   81 - 90 | *
   91 - 100| *

Answer 1

是的，counter[(list.get(i) - 1) / 10]++;或类似的东西应该能够替换所有那些if语句。

编辑：根据您更改的要求，您需要首先测试索引是否超出范围：

int index = (list.get(i) - 1 ) / 10;
if (index >= 0 || index < counter.length) {
    counter[index]++;
} else {
    // not sure if you need to do something for index out of bounds
}

Answer 2

在收集大小为10的段中的值时，可以使用/（整数除法）来计算索引：

for (int i=0; i<list.size(); i++) {
    int value = Math.min(list.get(i), 100);
    int index = value / 10 - 1;
    counter[index]++;
}

Answer 3

counter[(list.get(i) - 1) / 10]++;似乎最简洁。您还可以在for循环中使用switch statement来删除ifs并将所有内容放在同一代码块中。此外，将元素存储在第i个索引中可以节省大量重复get ting

编辑：原始代码与Java语法不一致。这是更新版本。 switch中的表达式是关键：

for(int i = 0; i < lst.size(); i++){
  int x = lst.get(i);
  switch((x-1)/10){
    case 1: counter[0]++;
    case 2: counter[1]++;
    case 3: counter[2]++;
    case 4: counter[3]++;
    case 5: counter[4]++;
    case 6: counter[5]++;
    case 7: counter[6]++;
    case 8: counter[7]++;
    case 9: counter[8]++;
  }
}

histogram：if语句和格式的快捷方式

3 个答案: