Question

我想使用字典系统在文本中打印每个字符的出现情况：

例如：文字为“我喜欢苹果”

控制台输出如下：

'i'在以下位置出现2次：1、4

'l'在以下位置出现2次：3、11

..

到目前为止，我已经拥有了

String text = "text";
HashMap<Integer, String> dictionary = new HashMap<Integer, String>();

for (int i = 0; i < text.length(); i++) {
    dictionary.put(i, String.valueOf(text.charAt(i)));
}

基本上只是将每个字母添加到字典的键值中，但我不知道如何进行打印...

Answer 1

这就是我要做的事情：

String[] tempArray = data.split("");
IntStream.rangeClosed(1, tempArray.length)            
         .boxed()
         .collect(groupingBy(index -> tempArray[index-1].toUpperCase()))
         .forEach((k, v) -> System.out.println(k + " has an occurrence of " +
                     v.size() + " times on positions " + v.stream()
                    .map(Object::toString).collect(joining(","))));

首先将字符串分成代表字符串的各个字符
利用IntStream.range生成索引
利用groupingBy对各个字符进行分组，并将其值作为List<Integer>来表示字符出现在其中的索引
最后，使用forEach格式化数据并打印到控制台。

给定的数据设置如下：

String data = "I like apples";

这将产生输出：

P has an occurrence of 2 times on positions 9,10
  has an occurrence of 2 times on positions 2,7
A has an occurrence of 1 times on positions 8
S has an occurrence of 1 times on positions 13
E has an occurrence of 2 times on positions 6,12
I has an occurrence of 2 times on positions 1,4
K has an occurrence of 1 times on positions 5
L has an occurrence of 2 times on positions 3,11

请注意，上述解决方案也将空格视为字符，如果您不希望在输出中使用空格，请使用filter操作将其排除：

IntStream.rangeClosed(1, tempArray.length)
         .filter(index -> !tempArray[index-1].trim().isEmpty())
         .boxed()
         .collect(groupingBy(index -> tempArray[index-1].toUpperCase()))
         .forEach((k, v) -> System.out.println(k + " has an occurrence of " +
                     v.size() + " times on positions " + v.stream()
                    .map(Object::toString).collect(joining(","))));

Answer 2

此代码使用字典，并以要求的确切格式打印正确答案：

打印到控制台：

import java.util.ArrayList;
import java.util.HashMap;

public class TestDriver {

public static void main(String[] args) {    

    String text = "i like apples";  
    char[] textArray = text.toCharArray();

    //a dictionary that will hold the letter as the key and a list of it's positions as the value.
    HashMap<Character, ArrayList<Integer>> dictionary = new HashMap<Character, ArrayList<Integer>>();

    //loop through the text to check each letter
    for (int i = 0; i < textArray.length; i++) {            

        //if you've already checked this letter, skip to the next one
        if(dictionary.containsKey(textArray[i])) {
            continue;
        }       

        //add the letter's position to its position list
        ArrayList<Integer> positionList = new ArrayList<>();
        positionList.add(i);

        //compare the remaining letters in the text to the current letter
        for (int j = i+1; j < textArray.length; j++) {

                //if a letter matches, add it's position to the list
                if(textArray[i] == textArray[j]) {
                positionList.add(j);
            }   
        }               

        //add the letter and its list of positions to the dictionary
        dictionary.put(textArray[i], positionList);
    }

    //format the response
    for(char key : dictionary.keySet()) {
        ArrayList<Integer> positions = new ArrayList<>();
        positions = dictionary.get(key);
        System.out.println(key + " has an occurance of " + positions.size() + " on positions: " + positions);           
        }
    }
}

Answer 3

您可以尝试如下操作

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class GenericPlayAround {
    public static void main(String[] args) throws InterruptedException {
        String text = "I like apples";
        Map<Character, List<Integer>> dictionary = new HashMap<Character, List<Integer>>();

        for (int i = 1; i <= text.length(); i++) {
            Character c = text.charAt(i-1);
            List<Integer> list;
            if (dictionary.containsKey(c)) {
                list = dictionary.get(c);
            }
            else {
                list = new ArrayList<Integer>();
                dictionary.put(c, list);

            }
            list.add(i);
        }

        System.out.println(dictionary);
    }
}

输出为

{ =[2, 7], p=[9, 10], a=[8], s=[13], e=[6, 12], I=[1], i=[4], k=[5], l=[3, 11]}

从字典Java打印

3 个答案: