Question

我必须使用数组来计算字符的频率

public class abc {
    public static void main ( String[] args) {
        Scanner input = new Scanner (System.in);

        String Text;
        System.out.print("Enter a text string:");
        Text=input.nextLine();
        System.out.println("Input String:");
        System.out.println(Text);
        }
    }
}

Answer 1

根据你的所作所为，你有一个woking代码。主要有3个步骤：

字符数

我使用了HashMap映射字符来计算而不是数组。递增计数时，您必须首先检查角色是否已经在地图中。

以最大的数量排序

我使用具有相反顺序的SortedMap，将计数映射到具有此计数的字符列表。由于以前的地图，我填充它。注意：这个地图可能是首先创建的，但我认为它更容易理解，并且更接近原始代码。

打印前10个频率的字母

最后，只需浏览有序地图即可显示计数最多的字母。

希望它可以帮到你，如果你不理解某些部分，就提问。

public static void main(final String[] args){
    final Scanner input = new Scanner(System.in);

    String strInText;
    // final int intLetterCount[] = new int[52];
    final Map<Character, Integer> intLetterCount = new HashMap<>();
    System.out.print("Enter a text string:");
    strInText = input.nextLine();
    strInText = strInText.trim();
    System.out.println("");
    System.out.println("Input String:");
    System.out.println("   \"" + strInText + "\"");

    for (int i = 0; i < strInText.length(); i++)
    {
        final char character = strInText.charAt(i);
        final int value = character;
        if ((value >= 65 && value <= 90) || (value >= 97 && value <= 122))
        {
            // current character count
            final Integer currentCount = intLetterCount.get(character);

            // case the character was not already present : add it to the map with count 1
            if (currentCount == null)
            {
                intLetterCount.put(character, 1);
            }
            // case the character was present. Increment count
            else
            {
                intLetterCount.put(character, currentCount + 1);
            }
        }
    }

    /*
     * Sort by character usage
     */
    // create sorted map (useful to sort easily) with reverse order to have biggest count first.
    final SortedMap<Integer, List<Character>> sortedMap = new TreeMap<>(Collections.reverseOrder());

    // for all character found, add
    for (final Map.Entry<Character, Integer> entry : intLetterCount.entrySet())
    {
        // declare current count and current char
        final int currentCount = entry.getValue();
        final Character currentChar = entry.getKey();

        // get actual list of characters with the current count.
        List<Character> charactersWithCount = sortedMap.get(currentCount);

        // if the list is null, there was no char with the current count
        if (charactersWithCount == null)
        {
            // create list, add the current char to it, put it in map
            charactersWithCount = new ArrayList<>();
            charactersWithCount.add(currentChar);
            sortedMap.put(currentCount, charactersWithCount);
        }
        // else there was other characters with the same count.
        else
        {
            // just add the new character to the list.
            charactersWithCount.add(currentChar);
        }
    }

    System.out.println("Top 10 letter counts(Number) : ");
    /*
     * print the first 10 values with higher count
     */
    final int max = 10;
    int charPrinted = 0;
    // loop through the map sorted by count
    mainLoop: for (final Map.Entry<Integer, List<Character>> entry : sortedMap.entrySet())
    {
        // loop through the list of character with the current count
        for (final Character ch : entry.getValue())
        {
            // if we have already printed more than 10 characters, break out of the main loop.
            if (charPrinted >= max)
            {
                break mainLoop;
            }

            // print char
            System.out.println(ch + " : " + entry.getKey());
            charPrinted++; // increase printed char count.
        }
    }
}

Answer 2

应该有点像这样：

System.out.println("Top 10 letter counts(Number):");
for( int i=0; i<intLetterCount.length; i++){
   char curr = 'a'+i;
   System.out.print(curr+"-"+intLetterCount[i]+",");
}

Answer 3

您可以尝试将字符及其计数添加到哈希图，以便您可以像键值对一样将其打印

Answer 4

您还可以使用流API：

final String input = "I am an Unicorn";
final Map<Character, Long> collect =
        input.chars( ).mapToObj( c -> (char) c )
             .collect( Collectors.groupingBy( Function.identity( ), Collectors.counting( ) ) );

collect.forEach( ( character, count ) -> System.out.println( character + "-" + count ) );

或显示前10名：

collect.entrySet( ).stream( )
       .sorted( Comparator.<Entry<Character, Long>>comparingLong( Entry::getValue ).reversed() )
       .limit( 10 )
       .forEach( ( e ) -> System.out.println( e.getKey()+ "-" + e.getValue()) );

如何显示前十个常用字符

4 个答案: