Question

我写了一个类，通过将两位数相加来找到一个整数的总和。例如，0, 5, 12, 213, 999 and 1111的数字总数是分别为0,5,3,6,27和4。我需要写一个方法：

public static double digitTotalFrequency(int max, int digitTotal)

此方法应该return具有0,1 ... max范围内所有数字的给定digitTotal的数字的比例。例如，digitTotalFrequency(9, 3)为0.1，因为前10个数字中只有一个数字的总数为3.此外，digitTotalFrequency(99,3)为0.04（因为前100个数字中的数字）只有数字3,12,21和30的数字总数为30）。我很难做到这一点，请帮助，谢谢。

public static double digitTotalFrequency(int max, int digitTotal){ 
    int number = max; 
     while (number>0) { 
       digitTotal = digitTotal + (number%10); 
       number = number/10; 
     } 
  return digitTotal/max; 
}

Answer 1

您需要将任务分为两部分

保持0到最大范围内所有数字的总数的运行计数
保持每个单独数字的总数0到最大

这被称为动态编程或记忆。

尝试稍微考虑一下。如果你挣扎，我可以开始使用一些代码

编辑：程序代码

尝试阅读本文，看看到底发生了什么

public class Jnk {

    public static void main(String[] args) {

        System.out.println(digitTotalFrequency(99,3));
    }

    public static double digitTotalFrequency(int max, int digitTotal){ 

        HashMap<String,Integer> counts = new HashMap<String,Integer>();

        counts.put("0", 0);
        counts.put("1", 1);
        counts.put("2", 2);
        counts.put("3", 3);
        counts.put("4", 4);
        counts.put("5", 5);
        counts.put("6", 6);
        counts.put("7", 7);
        counts.put("8", 8);
        counts.put("9", 9);

        int [] runs = new int[1];

        if(max<=10)
            return (double)digitTotal/max;

        else
            getTotal(0,max,counts,runs,digitTotal);

        return (double)runs[0]/max; 
    }

    private static void getTotal(int currentNumber, int max, HashMap<String, Integer> counts,int [] runs, int target) {


        if(currentNumber==max)
            return;

        String key = getKey(currentNumber);

        if(counts.containsKey(key))
        {
            int total = counts.get(key);

            if (total==target)
            {
                System.out.println("Found "+currentNumber);
                runs[0] += 1;
            }
            getTotal(currentNumber+1,max,counts,runs,target);
        }
        else
        {   
            int total = getDigitCount(currentNumber);

            counts.put(key, total);

            if (total==target)
            {
                System.out.println("Found "+currentNumber);
                runs[0] += 1;
            }
            getTotal(currentNumber+1,max,counts,runs,target);
        }
    }

    private static Integer getDigitCount(int currentNumber) {

        int digitTotal = 0;
        while (currentNumber>0) { 
               digitTotal = digitTotal + (currentNumber%10); 
               currentNumber = currentNumber/10; 
             } 

        return digitTotal;
    }

    private static String getKey(int min) {

        String number = String.valueOf(min);

        char[] numberArray = number.toCharArray();
        Arrays.sort(numberArray);
        String key = String.valueOf(numberArray);

        return key;
    } 
}

一种计算数字频率的方法

1 个答案: