用其全貌替换缩写/俚语

Question

I am using a HashMap to store the full forms for abbreviations.

public class Test {
    public static void main(String[] args) {
        Map<String, String> slangs = new HashMap<String, String>();
        slangs.put("lol", "laugh out loud");
        slangs.put("r", " are ");
        slangs.put("n", " and ");
        slangs.put("idk", " I don't know ");
        slangs.put("u", " you ");
        Set set = slangs.entrySet();
        Iterator i = set.iterator();

        String sentence = "lol how are you";
        StringBuilder sb = new StringBuilder();

        for (String word : sentence.split(" ")) {
            while(i.hasNext()) {
                Map.Entry<String, String> me = (Map.Entry)i.next();
                if (word.equalsIgnoreCase(me.getKey())) {
                    sb.append(me.getValue());
                    continue;
                }
                sb.append(word);
            }
        }
        System.out.println(sb.toString());
    }
}

The Output is:

lollollollaugh out loudlol

What is wrong here and how do I solve it?

Answer 1

You are not supposed to iterate over the entries to find a match, you are supposed to use get(Object key) or getOrDefault(Object key, V defaultValue) to get the full form of a given abbreviation, otherwise instead of getting your full form with a time complexity of O(1) , you will get it with a O(n) which is of course not good in term of performances, you would lose the real benefit of having your key/value pairs in a Map . If you did it because of the case, simply put your keys only in lower case in your map and call get or getOrDefault with the word in lower case as below:

So your loop should be something like:

for (String word : sentence.split(" ")) {
    // Get the full form of the value of word in lower case otherwise use
    // the word itself
    sb.append(slangs.getOrDefault(word.toLowerCase(), String.format(" %s", word)));
}

Output:

laugh out loud how are you

---

Using the Stream API , it could simply be:

String result = Pattern.compile(" ")
    .splitAsStream(sentence)
    .map(word -> slangs.getOrDefault(word.toLowerCase(), word))
    .collect(Collectors.joining(" "));

Answer 2

Don't loop over the keys in the dictionary. Instead, just check whether the key is in the map and get the corresponding value. Also, don't forget to add the spaces back into the combined sentence.

for (String word : sentence.split(" ")) {
    if (slangs.containsKey(word.toLowerCase())) {
        sb.append(slangs.get(word.toLowerCase()));
    } else {
        sb.append(word);
    }
    sb.append(" ");
}

If you are using Java 8, you can also use String.join , Map.getOrDefault and Streams:

String s = String.join(" ", Stream.of(sentence.split(" "))
        .map(word -> slangs.getOrDefault(word.toLowerCase(), word))
        .toArray(n -> new String[n]));

This latter approach also has the benefit of not adding a space before the first or after the last word in the sentence.

Answer 3

Simply, I think you just need to check if slangs contain this keyword or not. Please check my code.

 public class Test {
    public static void main(String[] args) {

      Map<String, String> slangs = new HashMap<String, String>();
      slangs.put("lol", "laugh out loud");
      slangs.put("r", " are ");
      slangs.put("n", " and ");
      slangs.put("idk", " I don't know ");
      slangs.put("u", " you ");

      String sentence = "lol how are you";
      String[] words = sentence.split(" ");

      for (String word : words) {
        String normalizeWord = word.trim().toLowerCase();
        if(slangs.containsKey(normalizeWord)) {
            sentence = sentence.replace(word, slangs.get(normalizeWord));
        }
    }
    System.out.println(sentence);
  }
}