Question

我的代码是：

public class Main{

    public static void main(String[] args){
        WordGroup wordgroupOne= new WordGroup ("You can discover more about a person in an hour of play than in a year of conversation");
        WordGroup wordgroupTwo= new WordGroup ( "When you play play hard when you work dont play at all");

        String[] quoteOne = wordgroupOne.getWordArray();    
        String[] quoteTwo = wordgroupTwo.getWordArray();

        for (String words : quoteOne){
            System.out.println(words);
        }

        for (String words : quoteTwo){                     
            System.out.println(words);
        }

    }

}

WordGroup类：

import java.util.HashSet;
import java.util.HashMap;

public class WordGroup {
    public String words;

    public WordGroup (String getWords){
        words = getWords.toLowerCase();
    }

    public String[] getWordArray(){
        return words.split(" ");   
    }

    public HashSet<String> getWordSet(){
        HashSet<String> set = new HashSet<String>();
        String[] p = getWordArray();
        for (String items : p){
            set.add(items);
        }
        System.out.println(set);
        return set;
    }

    public HashMap<String, Integer> getWordCounts() {
        HashMap<String, Integer> map = new HashMap<String, Integer>();
        String[] q = getWordArray();
        for (String stuff : q) {
            Integer oldVal = map.get(stuff);
            if (oldVal == null){
                oldVal = 0;
            }
            map.put(stuff, oldVal+1);
        }
        System.out.println(map);
        return map;
    }

}

我要做的是使用两个WordGroups和的getWordSet（）方法迭代或循环返回返回的HashSet并从中打印出单词。

在两个WordGroups上调用getWordCounts（）。使用keySet（）来检索键集。循环遍历此集并打印出两个WordGroup的单词及其计数。

使用getWordSet（）方法从两个WordGroups中完成所有单词的完整集合。遍历新的HashSet以打印包含每个哈希映射的总计数的所有单词的完整列表。

我正在努力解决所有这些问题。非常感谢任何帮助!!

Answer 1

如果要创建组合列表或集合，则必须将列表和地图合并在一起。我把这个练习留给你。

 public static void main(String[] args)
 {

    WordGroup wg1 = new WordGroup(
            "You can discover more about a person in an hour of play than in a year of conversation");
    WordGroup wg2 = new WordGroup(
            "When you play play hard when you work dont play at all");

    wg1.processWord();

    // iterate through all the distinct words
    Set<String> dw1 = wg1.getDistinctWords();
    for (String s : dw1)
    {
        System.out.println(s);
    }

    // use map entry to iterate through the entry set
    Map<String, Integer> wc1 = wg1.getWordCounts();
    for (Map.Entry<String, Integer> entry : wc1.entrySet())
    {
        if (entry != null)
        {
            // use stringbuilder to build a temp string
            // instead of using +
            StringBuilder sb = new StringBuilder();
            sb.append(entry.getKey());
            sb.append(": ");
            sb.append(entry.getValue());

            System.out.println(sb);
        }
    }

}

 public class WordGroup
 {
// as a class, made the results of the process private
private String originalWord;
// we declare generic versions of the Collections, instead of the specific
// implementation
private Set<String> distinctWords;
private Map<String, Integer> wordCounts;

public WordGroup(String s)
{
    this.originalWord = s;
    // here we declare and initialize the specific implementation
    this.distinctWords = new HashSet<String>();
    this.wordCounts = new HashMap<String, Integer>();
}

public void processWord()
{
    List<String> toProcess = getWordList();
    if (toProcess != null && !toProcess.isEmpty())
    {
        for (String s : toProcess)
        {
            // the set will automatically figure out if it should be in the
            // set or not.
            this.distinctWords.add(s);
            // call the update or insert method
            upsertString(s);
        }
    }
}

// this splits the string into a list
// you could probably use a utility class from guava or something to do this
// but i have coded a naive version
private List<String> getWordList()
{
    List<String> splitList = new ArrayList<String>();

    // check to see if there is anything there
    if (this.originalWord != null && !this.originalWord.isEmpty())
    {
        String lowered = this.originalWord.toLowerCase();

        String[] splits = lowered.split(" ");
        if (splits != null)
        {
            int iSize = splits.length;
            if (iSize > 0)
            {
                // basically create a string
                for (int i = 0; i < iSize; i++)
                {
                    splitList.add(splits[i]);
                }
            }
        }
    }

    return splitList;
}

// helper method to see if we need to add to the count
private void upsertString(String s)
{
    if (s != null && !s.isEmpty())
    {
        if (this.wordCounts != null)
        {
            // default to 1, if its an insert
            Integer newCount = 1;

            // if it already exists we want to update
            if (this.wordCounts.containsKey(s))
            {
                Integer currentCount = this.wordCounts.get(s);
                if (currentCount != null)
                {
                    // update the count by 1
                    newCount += currentCount;
                }
            }

            // insert the new item
            // or overwrite, because it is the same key to the new count
            this.wordCounts.put(s, newCount);
        }
    }
}

public String getOriginalWord()
{
    return this.originalWord;
}

public void setOriginalWord(String originalWord)
{
    this.originalWord = originalWord;
}

public Set<String> getDistinctWords()
{
    return this.distinctWords;
}

public void setDistinctWords(Set<String> distinctWords)
{
    this.distinctWords = distinctWords;
}

public Map<String, Integer> getWordCounts()
{
    return this.wordCounts;
}

public void setWordCounts(Map<String, Integer> wordCounts)
{
    this.wordCounts = wordCounts;
}

}

从另一个字符串中的一个类调用方法（hashmap / hashset）。

1 个答案: