請設計一個算法,給一個字符串進行二進制編碼,使得編碼后字符串的長度最短。
輸入描述:
每組數據一行,為待編碼的字符串。保證字符串長度小於等於1000。
輸出描述:
一行輸出最短的編碼后長度。
輸入例子:
MT-TECH-TEAM
輸出例子:
33
import java.util.Comparator;
import java.util.HashMap;
import java.util.Map.Entry;
import java.util.PriorityQueue;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (scanner.hasNext()) {
String s = scanner.next();
System.out.println(createHuffman(s));
}
}
public static int createHuffman(String s) {
char[] chars = s.toCharArray();
//hash表存放每個字符和出現的次數(出現的次數就是權重)
HashMap<Character, Integer> map = new HashMap<Character, Integer>();
for(int i=0;i<chars.length;i++){
if(map.containsKey(chars[i])){
map.put(chars[i], map.get(chars[i]) + 1);
}
else{
map.put(chars[i], 1);
}
}
//優先隊列(最小推),每次能得到weigh最小的node
PriorityQueue<HuffmanNode> queue = new PriorityQueue<>(map.size(), new Comparator<HuffmanNode>() {
@Override
public int compare(HuffmanNode o1, HuffmanNode o2) {
// TODO Auto-generated method stub
return o1.weight - o2.weight;
}
});
//構建哈夫曼樹合並兩個權重最小的節點,直到只剩下根節點root
for(Entry<Character, Integer> entry : map.entrySet()){
queue.offer(new HuffmanNode(entry.getValue(), entry.getKey()));
}
while(queue.size() > 1){
//彈出兩個權重最小的,合並為一個node
HuffmanNode leftNode = queue.poll();
HuffmanNode rightNode = queue.poll();
HuffmanNode fatherNode = new HuffmanNode(leftNode.weight + rightNode.weight);
fatherNode.left = leftNode;
fatherNode.right = rightNode;
queue.offer(fatherNode);
}
HuffmanNode root = queue.poll();
return getLength(root, 0);
}
public static int getLength(HuffmanNode node, int depth) {
if(node == null){ //僅計算ch有值的
return 0;
}
return (node.ch == null ? 0:node.weight)*depth + getLength(node.left, depth+1) + getLength(node.right, depth+1);
}
}
class HuffmanNode{
int weight;//權重(字符出現次數)
HuffmanNode left;
HuffmanNode right;
Character ch; //如果是初始字符,則ch為字符,如果是合並的,則為null
public HuffmanNode(int weight) {
super();
this.weight = weight;
}
public HuffmanNode(int weight, char ch) {
super();
this.weight = weight;
this.ch = ch;
}
}
