java如何在数字字符串中查找最频繁的数字？

Question

/*我是编程新手。作为作业的一部分，我必须找到数字字符串中最常见的数字。虽然我的代码被编译了，但它并没有给出正确的结果。请帮我做这个。 */

    import java.util.*;

    public class program
    {
    public int test(String number)
    {
    int freq0,freq1,freq2,freq3,freq4,freq5,freq6,freq7,freq8,freq9;
    freq0=0;
    freq1=0;
    freq2=0;
    freq3=0;
    freq4=0;
    freq5=0;
    freq6=0;
    freq7=0;
    freq8=0;
    freq9=0;
    for (int i =0; i<number.length();i++)
    {
        switch (number.charAt(i))
        {
            case 48: freq0++;
            break;
            case 49: freq1++;
            break;
            case 50: freq2++;
            break;
            case 51: freq3++;
            break;
            case 52: freq4++;
            break;
            case 53: freq5++;
            break;
            case 54: freq6++;
            break;
            case 55: freq7++;
            break;
            case 56: freq8++;
            break;
            case 57: freq9++;
            break;
        }
    }
    List<Integer> hope = new ArrayList<Integer>();

    hope.add(freq0);
    hope.add(freq1);
    hope.add(freq2);
    hope.add(freq3);
    hope.add(freq4);
    hope.add(freq5);
    hope.add(freq6);
    hope.add(freq7);
    hope.add(freq8);
    hope.add(freq9);

    int temp=0;
    for (int j=0; j<(hope.size());j++)
    {
        if (temp<hope.get(j))
        {
            temp=hope.get(j);
        }
    }

    return temp;
    }
    }

//该类由另一个主类调用。我也在写代码

    public class checker
    {
    public static void main ( String args [])
    {
    int inputs[] = {1234, 11, 144, 97764 };
    int outputs[] = {1, 1, 4, 7};

    for(int i=0; i<inputs.length; i++) {
        int input=inputs[i];
        int oracle_output=outputs[i];

        program p = new program();

        String input_string = "" + input;
        int output = p.test(input_string);
        if(output==oracle_output) {
            System.out.println("test passed for " + input);
        } else {
            System.out.println("test failed for " + input);
        }
    }
    }
    }

Answer 1

您所能做的就是使用一个简单的数组

long number = 12345612;
int []frequency = new int[10];
while(number > 0)
{
 int digit = number % 10;
 number /= 10;
 frequency[digit] ++;
}

for(int i = 0; i < 10; ++i)
 System.out.println(frequency[i]);

---

如果它是一个String

String str = "12342352397235823050237238523";
int []frequency = new int[10];
for(int i = 0; i < str.length(); ++i)
{
 int digit = str.charAt(i) - '0';
 frequency[digit] ++;
}
for(int i = 0; i < 10; ++i)
 System.out.println(frequency[i]);

以获取最大占用位数

int maxFrequency = 0;
int index = 0;
for(int i = 0; i < 10; ++i){
   if(frequency[i] > maxFrequency){
     maxFrequency = frequency[i];
     index = i;
    }
  }
System.out.println("The highest occuring digit is " + index + " occuring " + maxFrequency + " times(s)");

你程序中的错误是你没有考虑重复多次的数字，你只考虑了重复的次数。因此，你只得到频率，而不是数字

Answer 2

以下是可读性更强、更简单的代码：

            import java.util.ArrayList;
            import java.util.List;

            public class MostFrequentDigits1 {

                /**
                 * @param args
                 */
                public static void main(String[] args) {
                    // TODO Auto-generated method stub

                    long number = 12345612;
                    String str = "12342352397235823050237238523";
                    getMostFrequentDigitWInNumber(number);
                    getMostFrequentDigitInString(str);

                }

                public static void getMostFrequentDigitWInNumber(long number) {
                    List<Integer> list = new ArrayList<Integer>();
                    int[] frequency = new int[10];
                    int mostFrequentVal = 0;
                    while (number > 0) {
                        int digit = (int) (number % 10);
                        number /= 10;
                        frequency[digit]++;
                        if (frequency[digit] > mostFrequentVal) {
                            mostFrequentVal = frequency[digit];
                        }
                    }

                    for (int j = 0; j < frequency.length; j++) {
                        if (frequency[j] == mostFrequentVal) {
                            list.add(j);
                        }
                    }
                    System.out.println("most frequent digits in Number:" + list);
                }

                public static void getMostFrequentDigitInString(String str) {
                    List<Integer> list = new ArrayList<Integer>();
                    int mostFrequentVal = 0;
                    int[] frequency = new int[10];
                    for (int k = 0; k < str.length(); k++) {
                        char c = str.charAt(k);
                        int ind = Integer.valueOf(String.valueOf(c));
                        frequency[ind]++;
                        if (frequency[ind] > mostFrequentVal) {
                            mostFrequentVal = frequency[ind];
                        }
                    }

                    for (int j = 0; j < frequency.length; j++) {
                        if (frequency[j] == mostFrequentVal) {

                            list.add(j);
                        }
                    }
                    System.out.println("most frequent digits in string:" + list);
                }

            }

Answer 3

正如注释中所说，代码可以变得更简单（更可读） 但是，您的问题的答案是：返回值（temp）是最频繁数字的频率，而不是数字本身（即j，您需要“记住”它）