编写程序以打印每个字母以及用户输入中出现了多少个字母

Question

I have to write a program to accept a String as input, and as output I'll have to print each and every alphabetical letter, and how many times each occurred in the user input. There are some constraints:

• I cannot use built-in functions and collection
• The printed result should be sorted by occurrence-value.

For example, with this input:

abbbccccdddddzz

I would expect this output:

a-1,z-2,b-3,c-4,d-5

This is what I have so far:

public static void isCountChar(String s) {
    char c1[] = s.toCharArray();
    int c3[] = new int[26];
    for (int i = 0; i < c1.length; i++) {
        char c = s.charAt(i);
        c3[c - 'a']++;
    }

    for (int j = 0; j < c3.length; j++) {
        if (c3[j] != 0) {
            char c = (char) (j + 'a');
            System.out.println("character is:" + c + " " + "count is:  " + c3[j]);
        }
    }
}

But I don't know how to sort.

Answer 1

some sort: easy if not easiest to understand an coding

• You loop from the first element to the end -1 : element K

• compare element K and element K+1: if element K>element K+1, invert them

• continue loop

• if you made one change redo that !

Answer 2

First of all a tip for your next question: The things you've stated in your comments would fit better as an edit to your question. Try to clearly state what the current result is, and what the expected result should be.

That being said, it was an interesting problem, because of the two constraints.

• First of all you weren't allowed to use libraries or collections. If this wasn't a constraint I would have suggested a HashMap with character as keys, and int as values, and then the sorting would be easy.
• Second constraint was to order by value. Most people here suggested a sorting like BubbleSort which I agree with, but it wouldn't work with your current code because it would sort by alphabetic character instead of output value .

With these two constraints it is probably best to fake key-value pairing yourself by making both an keys -array and values -array, and sort them both at the same time (with something like a BubbleSort -algorithm). Here is the code:

private static final int ALPHABET_SIZE = 26;

public static void isCountChar(String s)
{
    // Convert input String to char-array (and uppercase to lowercase)
    char[] array = s.toLowerCase().toCharArray();

    // Fill the keys-array with the alphabet
    char[] keys = new char[ALPHABET_SIZE];
    for (int i = 0; i < ALPHABET_SIZE; i++)
    {
        keys[i] = (char)('a' + i);
    }

    // Count how much each char occurs in the input String
    int[] values = new int[ALPHABET_SIZE];
    for (char c : array)
    {
        values[c - 'a']++;
    }

    // Sort both the keys and values so the indexes stay the same
    bubbleSort(keys, values);

    // Print the output:
    for (int j = 0; j < ALPHABET_SIZE; j++)
    {
        if (values[j] != 0)
        {
            System.out.println("character is: " + keys[j] + "; count is: " + values[j]);
        }
    }
}

private static void bubbleSort(char[] keys, int[] values)
{
    // BUBBLESORT (copied from http://www.java-examples.com/java-bubble-sort-example and modified)
    int n = values.length;

    for(int i = 0; i < n; i++){
        for(int j = 1; j < (n - i); j++){        
            if(values[j-1] > values[j]){
                // Swap the elements:
                int tempValue = values[j - 1];
                values[j - 1] = values[j];
                values[j] = tempValue;

                char tempKey = keys[j - 1];
                keys[j - 1] = keys[j];
                keys[j] = tempKey;
            }         
        }
    }
}

Example usage:

public static void main (String[] args) throws java.lang.Exception
{
    isCountChar("TestString");
}

Output:

character is: e; count is: 1
character is: g; count is: 1
character is: i; count is: 1
character is: n; count is: 1
character is: r; count is: 1
character is: s; count is: 2
character is: t; count is: 3

Here is a working ideone to see the input and output.