java - 破解面试19.8-创建哈希表时，当我在 if-else 语句中添加 else 时，为什么会得到错误的输出

Question

破解面试中19.8的问题是:设计一种方法来查找任意给定单词在书中出现的频率。

书中提供的解决方案如下所示，除了我在 createHashtable 函数中添加了 else 之外。

我不知道为什么当我输入时会得到错误的输出(输出是:1, 0, 0，应该是2, 1, 1) if 后面的 else 语句。

import java.util.Hashtable;

public class Question198
{
  /*create a hashtable for the strings in the book*/
  public static Hashtable<String, Integer> createHashtable(String[] book) {
    Hashtable<String, Integer> table = new Hashtable<String, Integer>();
    for(int i = 0; i < book.length; i++) {
      book[i] = book[i].toLowerCase();
      if(book[i].trim() != " ") {
        if( !table.containsKey(book[i])) {
          table.put(book[i], 0);
        } 
        else{  // ?? why else is here is wrong??
          table.put(book[i], table.get(book[i]) + 1);
        }
      }
    }
    return table;
  }
   /*get the frequency of the given word in the book*/
  public static int getFrequency(Hashtable<String, Integer> table, String word) {
    if( table == null || word == null) return -1;
    word = word.toLowerCase();
    if( table.containsKey(word)) {
      return table.get(word);
    }
    return 0;
  }

/*Test*/
  public static void main(String[] args) {
    String[] book1 = {"This", "is", "a", "is", "case",};
    Hashtable<String, Integer> test1 = createHashtable(book1);
    String word1 = "is";
    String word2 = "a";
    String word3 = "case";
    System.out.println("Expected(2): " + getFrequency(test1, word1));
    System.out.println("Expected(1): " + getFrequency(test1, word2));
    System.out.println("Expected(1): " + getFrequency(test1, word3));
  }

}

Answer 1

您不需要输入 1 个初始出现的位置吗？

table.put(book[i], 1);

在代码中

if( !table.containsKey(book[i])) {
  table.put(book[i], 0);
} 

是的，trim() 在任何情况下都不会为您留下 ""