java - 这段代码有什么问题？霍夫曼编码

Question

在这段代码中，我试图将一个字符串拆分为多个字符，并将每个字符放入一个映射中。如果同一个角色出现不止一次，我会在其上放置一个计数器并将其放回 map 中，增加整数(频率)。

public class FrequencyMap {
   public static Map<Character, Integer> generateMap(String s){
       HashMap<Character, Integer> myMap = new HashMap<Character, Integer>();
       //generate a map of frequencies of the characters in s
       //need to break the string down into individual characters, sort them
       //in there frequencies then place them in map
       for(int i = 0; i < s.length(); i++){
           //break string into characters
           //need to determine how many characters make up the string, can do this by 
           //putting a counter on each letter that appears when the string is being 
           //broken down, if a letter reoccurs increment the counter by one.
           s.substring(i);
           char ch = s.charAt(i);
           myMap.put(ch, i);
           //calculating the occurence of a character in a string.
           if(myMap.containsKey(ch)){                   
               myMap.put(ch, myMap.get(i) + 1);                   
                  }//end of if statement
           }//end of for loop          
           return myMap;
       }//end of public generateMap()
   }//end of FrequencyMap

这是主要内容

   public static void main(String args[]){

       String str = "missisippi";

       Map frequencyMap = FrequencyMap.generateMap(str);
       HuffmanTree tree = new HuffmanTree(frequencyMap);
       Map<Character, String> encodingMap = tree.getEncodingMap();

       String encoded = tree.encode(str, encodingMap);
       System.out.println(encoded);     
   }//end of main  

Answer 1

好吧，有几件事...

字符串是不可变的!!

s.substring(i);

应该是

s = s.substring(i);

虽然我仍然不太清楚这到底是什么意思。

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其次..

这些行没有意义

myMap.put(ch, i);

if(myMap.containsKey(ch)){                   
    myMap.put(ch, myMap.get(i) + 1);                   
}

您刚刚添加了 key ch然后你立即询问 map 是否包含 ch - 这将永远是正确的。

我想你可能想把 if -statement first 并放置 myMap.put(ch, <b>1</b>)在else -条款。哦，还有myMap.get(i)可能应该是myMap.get(ch) .