Question

在这段代码中，我试图将字符串拆分为字符并将每个字符放入地图中。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到地图中，然后递增整数（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

这是主要的

   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

好的一些事情......

字符串是不可改变的!!

s.substring(i);

应该是

s = s.substring(i);

虽然我还不太确定这一点是什么意思。

其次..

这些行没有意义

myMap.put(ch, i);

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

您刚刚添加了密钥ch，然后您立即询问地图是否包含ch - 这将始终为真。

我认为您可能打算将if - 语句放在第一个并将myMap.put(ch, 1)放入else - 子句中。哦，myMap.get(i)可能应该是myMap.get(ch)。

Answer 2

您正在使用角色位置初始化您的计数器：

myMap.put(ch, i);

你想要的地方：

myMap.put(ch, 1);

您还需要在初始化计数器之前检查已存在的字符的地图，增加计数器的计数器（使用get(ch)而不是get(i)），给出：

char ch = s.charAt(i); //calculating the occurence of a character in a string. if (myMap.containsKey(ch)){ myMap.put(ch, myMap.get(ch) + 1); } else { myMap.put(ch, 1); }//end of if statement

这段代码有什么问题？霍夫曼编码

2 个答案: