我需要分析一个满足以下问题的文本：
生成一个输出，显示文本中每个单词出现次数的计数。报告应首先按字长排序，然后按自然排序。
我在下面给出了解决方案的停靠点。有更好的解决方案吗？我没有使用过地图或任何收藏品，因为我们被告知那些不使用收藏品的人将获得额外的信用。
我的POJO
/**
* An instance of this object represents the string that occurs in a sentence
* and the number of times it occurs in a single string.
*/
public class Word implements Comparable<Word> {
private final String word;
private int counter = 1;
public Word(String word) {
this.word = word;
}
public void incrementCounter() {
this.counter ++;
}
public String getWord() {
return word;
}
public int getCounter() {
return counter;
}
/**
* Overrides the default hashcode function.
*/
@Override
public int hashCode() {
int hashCode = 103034;
hashCode += this.word != null ? this.word.hashCode() ^ 3 : 0;
hashCode += this.counter ^ 2;
return hashCode;
}
/**
* Overrides the default equals function.
*/
@Override
public boolean equals(Object obj) {
if (obj == null) {
return false;
}
if (this == obj) {
return true;
}
if (!(obj instanceof Word)) {
return false;
}
Word otherWord = (Word) obj;
if (this.word != null && this.word.equals(otherWord.getWord())) {
return true;
}
return false;
}
@Override
public String toString() {
final StringBuilder sb = new StringBuilder();
sb.append("Word");
sb.append("{word='").append(word).append('\'');
sb.append(", counter=").append(counter);
sb.append('}');
return sb.toString();
}
/**
* The implementation checks for the order of comparison.
* The implementation first compares by presence of word string.
*
* @param w Word to be compared
* @return calculated order of comparison.
*/
@Override
public int compareTo(Word w) {
if (w == null) {
return -1;
}
if (this.word == null && w.getWord() != null) {
return 1;
}
if (this.word != null && w.getWord() == null) {
return -1;
}
return StringUtils.compareString(this.word, w.getWord());
}
}
import java.util.Comparator;
/**
* An instance of this class is responsible for comparing the instances of two word
* instances by comparing against the word string.
*/
public class WordComparator implements Comparator<Word> {
/**
* Compares two instances of words first by length of the word string and then by the
* word itself.
*
* @param firstWord first word to be compared
* @param secondWord second word to be compared
* @return negative number if the first word is less than the second word;
* positive if the first word is greater than the second word; 0 if equal.
*/
@Override
public int compare(Word firstWord, Word secondWord) {
if (firstWord == secondWord) {
return 0;
}
if (firstWord != null && secondWord == null) {
return -1;
}
if (firstWord == null && secondWord != null) {
return 1;
}
return StringUtils.compareString(firstWord.getWord(), secondWord.getWord());
}
}
实用工具类。
public class StringUtils {
// Not to be instantiated.
private StringUtils() {}
/**
* Compares the string first by word length and then by string.
*
* @param first First string to be compared.
* @param second Second string to be compared
* @return integer representing the output of comparison.
*/
public static int compareString(String first, String second) {
if (first == second) {
return 0;
}
if (first == null && second != null) {
return 1;
}
if (first != null && second == null) {
return -1;
}
int wordLengthDifference = first.length() - second.length();
if (wordLengthDifference == 0) {
return first.compareTo(second);
}
return wordLengthDifference;
}
}
主要方法：
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
/**
* Finds the number of occurrences of word a in given string. The implementation expects
* the input to be passed adsingle string argument.
*/
public class StringWordOccurences {
/**
* Finds the number of occurrences of a word in a string. The implementation relies
* on the following assumptions.
*
* <p>The word is passed as separate strings as in {@code "Hello" "World"} instead of
* a single string {@code "Hello World"}.
*
* @param args Arguments to be sorted by first word length and then by string.
*/
public static void main(String[] args) {
if (args == null || args.length == 0) {
System.out.println("There were no words. The count is 0");
return;
}
// Find the number of unique words and put them in an array.
Comparator<Word> wordComparator = new WordComparator();
Arrays.sort(args, new Comparator<String>() {
@Override
public int compare(String first, String second) {
return StringUtils.compareString(first, second);
}
});
Word [] words = new Word[args.length];
int numberOfUniqueWords = 0;
for (String wordAsString : args) {
Word word = new Word(wordAsString);
int index = Arrays.binarySearch(words, word, wordComparator);
if (index > -1) {
words[index].incrementCounter();
} else {
words[numberOfUniqueWords ++] = word;
}
}
Word [] filteredWords = Arrays.copyOf(words, numberOfUniqueWords);
// The display output.
for (Word word : filteredWords) {
System.out.println(word);
}
}
}
答案 0 :(得分：1)
在按所需顺序对单词进行排序后，您不需要二进制搜索，因为相同的单词将彼此相邻。只需翻阅所有单词并将每个单词与前一个单词进行比较。如果它们相等，则递增计数器。如果它们不相等，您可以立即打印完成的前一个单词的结果（无需将结果存储在数组中）。你也不需要Word类，你可以使用字符串。
答案 1 :(得分：0)
我能想到的几点，但显然这些是第二次猜测......