Java中的字符串频率计数器

Question

I have a string that a user will type in representing time. ie "10:05"

I've been searching for a while, and cannot find a way to search this string to find a specific number. The reason I need this number, is because I need to take each number and use an algorithm to discover how much power it takes to display this number on a digital clock. To do so, I need the frequency of each number. For example, if the time was 10:05, the number 0 would occur 2 times, and I can take that 2 and multiply it by the necessary numbers to discover the power needed.

Answer 1

I'm sure there's a more efficient way to do this, but:

String test = "10:05";

char[] array = test.toCharArray();
int counter = 0;

for(char c : array) {
   if (c.Equals('0')) {
       counter++;
   }
}

System.out.println("The number '0' appears " + counter + " times."); //2

Answer 2

Create a map containing the number you want as key and the frequency that each number that appears as value. Iterate through the string, character by character, disregarding non-digit characters increment the digit-frequency mapping as you go.

Example:

HashMap<Integer, Integer> num_freq = new HashMap<>();

String input = "10:05"; // Example input

char[] input_chars = input.toCharArray();

for(char c : input_chars){
    // Accept only characters that are digits
    if(Character.isDigit(c)){
        // Grabs digit from character

        int num = Character.digit(c, 10);

        // Put 1 into map if no entry exists, else increment existing value

        num_freq.compute(num, (k_num, freq) -> freq == null ? 1 : freq + 1);
    }
}

// Print result out

num_freq.forEach((num, freq) -> {
    System.out.println("Digit " + num + " appears " + freq + " time(s)");
});

Answer 3

If I got your question, then below code will work, I hope -

int[] powerArr   = {2,3,2,3,2,3,2,3,2,3};
String input = "10:05";
int powerConsumption = 0;

for(char c : input.replace(":", "").toCharArray()) {
    powerConsumption = powerConsumption + powerArr[Integer.parseInt(String.valueOf(c))];
}

powerConsumption : Total power consumption
powerArr : array of power to display 0, then power to display 1 ... power to display 9

If you are interested on number of occurance only then -

int[] counterArr = {0,0,0,0,0,0,0,0,0,0};
String input = "10:05";
int tmpVal = 0;

for(char c : input.replace(":", "").toCharArray()) {
    tmpVal = Integer.parseInt(String.valueOf(c));
    counterArr[tmpVal] = counterArr[tmpVal]+1;
}

First value of counterArr will represent occurance of 0, second one of 1 and so on..

Answer 4

This worked for me: (Assuming that time is entered always in xx:yy format)

 public static void main(String args[])
    {
        String time = "10:07"; //User Input
        time = time.replace(":", "");
        char digit[] = {time.charAt(0), time.charAt(1), time.charAt(2), time.charAt(3)};
        int[] count = new int[digit.length];
        Arrays.sort(digit);

        for (int i = 0; i < digit.length; i++)
        {
            count[i]++;
            if (i + 1 < digit.length)
            {
                if (digit[i] == digit[i + 1])
                {
                    count[i]++;
                    i++;
                }
            }
        }

        for (int i = 0; i < digit.length; i++)
        {
            if (count[i] > 0)
            {
                System.out.println(digit[i] + " appears " + count[i]+" time(s)");
            }
        }
    }

Output:

0 appears 2 time(s)
1 appears 1 time(s)
7 appears 1 time(s)