这里是代码...
public class Huffman_Coding {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.println("Enter a string to compress: ");
String str = sc.nextLine();
sc.close();
HashString hs = new HashString();
HashMap<Character, Integer> hm = hs.getStringHash(str);
PriorityQueue<Node> pq = new PriorityQueue<Node>();
for (char ch : hm.keySet()) {
pq.add(new Node(null, null, hm.get(ch), ch));
}
System.out.println(pq);
while (pq.size() != 1) {
Node left = pq.poll();
Node right = pq.poll();
Node parent = new Node(left, right, left.freq + right.freq, '\0');
pq.add(parent);
System.out.println(pq);
}
Huffman_Tree ht = new Huffman_Tree();
String ans = "";
ht.inOrder(pq.poll(), ans);
}
}
class Node implements Comparable<Node> {
@Override
public String toString() {
return "Node [freq=" + freq + ", ch=" + ch + "]";
}
Node lptr;
Node rptr;
int freq;
char ch;
Node(Node lptr, Node rptr, int freq, char ch) {
this.freq = freq;
this.lptr = lptr;
this.rptr = rptr;
this.ch = ch;
}
public int compareTo(Node o) {
int comparedvalue = Integer.compare(this.freq, o.freq);
if (comparedvalue != 0)
return comparedvalue;
else
return Integer.compare(this.ch, o.ch);
}
}
boolean isLeaf() {
return this.lptr == null && this.rptr == null;
}
}
class Huffman_Tree {
void inOrder(Node root, String code) {
if (!root.isLeaf()) {
inOrder(root.lptr, code + '0');
inOrder(root.rptr, code + '1');
} else
System.out.println(root.ch + " : " + code);
}
}
这里,对于输入字符串abccddeeee
,我得到类似的东西:
[Node [freq=1, ch=a], Node [freq=1, ch=b], Node [freq=2, ch=c], Node [freq=2, ch=d], Node [freq=4, ch=e]]
[Node [freq=2, ch= ]]
我很困惑,为什么在第二步中,具有'd'的节点要从'e'开始。这使我在最终编码中出错。为什么compareTo方法失败,我无法理解。
getHashString返回一个哈希,该哈希在键中包含字符,在值中包含其频率。
我不知道为什么在[[polling元素ant adding新的“ synthetic”元素之后PriorityQueue
中元素的顺序为什么不是期望的顺序,但是我认为您可以解决问题切换到TreeSet
,就像我用[]成功完成的TreeSet<Node> pq = new TreeSet<Node>((n1, n2) -> n1.compareTo(n2)); // explicit but unnecessary comparator
并将每个pq.poll()
调用更改为pq.pollFirst()
...
我希望此解决方法可以为您提供帮助!