Question

有人可以告诉我这个霍夫曼代码中的空间和时间复杂性，用Bog O符号来解释一下。非常感谢，谢谢。请分别提一下每种方法的Big O，会很棒。感谢。

package HuffmanProject;

import java.util.*;

class MyHCode {
public static void main(String[] args) {
String test = "My name is Zaryab Ali";
int[] FreqArray = new int[256];

for (char c : test.toCharArray()) {
FreqArray[c]++;
        }
MyHTree tree = ImplementTree(FreqArray);

System.out.println("CHARACTER\tFREQUENCY\tBINARY EQUIVALEENT CODE");
PrintMyHCode(tree, new StringBuffer());
    }

public static MyHTree ImplementTree(int[] FreqArray) {
PriorityQueue<MyHTree> trees = new PriorityQueue<MyHTree>();

for (int i = 0; i < FreqArray.length; i++) {
if (FreqArray[i] > 0) {
trees.offer(new MyHLeaf(FreqArray[i], (char) i));
            }
        }
while (trees.size() > 1) {
MyHTree FChild = trees.poll();
MyHTree SChild = trees.poll();
trees.offer(new MyHNode(FChild, SChild));
        }
return trees.poll();
    }

public static void PrintMyHCode(MyHTree tree, StringBuffer prefix) {

if (tree instanceof MyHLeaf) {
MyHLeaf leaf = (MyHLeaf) tree;
System.out.println(leaf.CharValue + "\t\t" + leaf.frequency + "\t\t" + prefix);
} 
else if (tree instanceof MyHNode) {
            MyHNode node = (MyHNode) tree;

            prefix.append('0');
            PrintMyHCode(node.left, prefix);
            prefix.deleteCharAt(prefix.length() - 1);

            prefix.append('1');
            PrintMyHCode(node.right, prefix);
            prefix.deleteCharAt(prefix.length() - 1);
        }
    }


}
abstract class MyHTree implements Comparable<MyHTree> {
public int frequency;

public MyHTree(int f) {
        frequency = f;
    }

public int compareTo(MyHTree tree) {
        return frequency - tree.frequency;
    }
}

class MyHLeaf extends MyHTree {
public char CharValue;

public MyHLeaf(int f, char v) {
        super(f);
        CharValue = v;
    }
}

class MyHNode extends MyHTree {
public MyHTree left, right;

public MyHNode(MyHTree l, MyHTree r) {
        super(l.frequency + r.frequency);
        left = l;
        right = r;
    }
}

Answer 1

PrintMyHCode（）方法遍历左侧和右侧。右子树，直到找到leaft节点。如果树中有n个元素，则此方法的复杂性为O（n）。

ImplementTree（）方法将数组中的值添加到树中，然后轮询其子项。如果数组中有n个元素： 1.此方法中for循环的复杂性将为O（n），因为每个元素都直接添加到树中 2.此方法中while循环的复杂性将为O（logn），假设每个节点至少有两个子节点。

因此，Big O表示法中ImplementTree（）方法的总时间复杂度为O（nlogn）。

希望，这个答案适合你。

我的霍夫曼算法的大O复杂性

1 个答案: