Trie实施-Java 学习之路

我试图在Java中实现一个非常简单的Trie，支持3个操作 . 我希望它有一个insert方法，一个has方法（即trie中的某个单词），以及一个以字符串形式返回trie的toString方法 . 我相信我的插入工作正常，但是并且toString被证明是困难的 . 这是我到目前为止所拥有的 .

特里班 .

public class CaseInsensitiveTrie implements SimpleTrie {

    //root node
    private TrieNode r;

    public CaseInsensitiveTrie() {
        r = new TrieNode();
    }

    public boolean has(String word) throws InvalidArgumentUosException {
        return r.has(word);
    }

    public void insert(String word) throws InvalidArgumentUosException {
        r.insert(word);
    }

    public String toString() {
        return r.toString();
    }

    public static void main(String[] args) {

        CaseInsensitiveTrie t = new CaseInsensitiveTrie();

        System.out.println("Testing some strings");
        t.insert("TEST");
        t.insert("TATTER");
        System.out.println(t.has("TEST"));
    }
}

和节点类

public class TrieNode {

    //make child nodes
    private TrieNode[] c;
    //flag for end of word
    private boolean flag = false;

    public TrieNode() {
        c = new TrieNode[26]; //1 for each letter in alphabet
    }

    protected void insert(String word) {
        int val = word.charAt(0) - 64;

        //if the value of the child node at val is null, make a new node
                //there to represent the letter
        if (c[val] == null) {
            c[val] = new TrieNode();
        }

        //if word length > 1, then word is not finished being added.
        //otherwise, set the flag to true so we know a word ends there.
        if (word.length() > 1) {
            c[val].insert(word.substring(1));
        } else {
            c[val].flag = true;
        }
    }

    public boolean has(String word) {
        int val = word.charAt(0) - 64;
        if (c[val]!=null && word.length()>1) {
            c[val].has(word.substring(1));
        } else if (c[val].flag==true && word.length()==1) {
            return true;
        }

        return false;
    }

    public String toString() { 
        return "";
    }
}

所以基本上，在创建Trie时，TrieNode被创建为具有26个子节点的根 . 尝试插入时，将在该根节点上调用insert，该节点以正确的位置递归创建新节点，并继续直到该单词完成 . 我相信这种方法运作正常 .

我的函数非常破碎，因为出于某种原因我必须在括号外面有返回语句 . 我不能在else子句中包含它或编译器抱怨 . 除此之外，我认为这种方法应该有一些调整，但我无法弄清楚我的生活 .

toString是一个我试图解决的野兽，但我没有投入任何东西，所以我会离开它，直到我解决了问题 . 如果我开始工作，我可能想办法将其重新格式化为toString函数 .

int val = word.charAt（0） - 64的目的;是因为输入的每个字符串必须全部大写（我将创建一个字符串格式化函数以确保此后）所以第一个字母的int值 - 64将是它在数组中的位置 . 即，数组索引0是A，所以A = 64，A = 64 = 0. B = 65，B = 64 = 1，依此类推 .

6 回答

您的 has 函数应该如下所示：

if (c[val]!=null && word.length()>1) {
    return c[val].has(word.substring(1)); //<-- Change is on this line
} else if (c[val].flag==true && word.length()==1) {
    ...etc

您执行递归调用，但您确实需要让该值传播回原始调用者 .

回复于 2024-04-29T09:12:29+08:00

1

也许您可以使用“Map c”而不是“TrieNode [] c”，这将允许您将此用于所有类型的字符大写/小写甚至特殊字符，甚至可以节省空间（在26处分配26个字符数组）每个角色级别）

回复于 2024-04-29T09:12:29+08:00

这是我的实施： -

public class Tries {

class Node {
    HashMap<Character, Node> children;
    boolean end;
    public Node(boolean b){
        children = new HashMap<Character, Tries.Node>();
        end = false;
    }
}
private Node root;
public Tries(){
    root = new Node(false);
}
public static void main(String args[]){
    Tries tr = new Tries();
    tr.add("dog");
    tr.add("doggy");

    System.out.println(tr.search("dogg"));
    System.out.println(tr.search("doggy"));
}
private boolean search(String word) {
    Node crawl = root;
    int n = word.length();
    for(int i=0;i<n;i++){
        char ch = word.charAt(i);
        if(crawl.children.get(ch) == null){
            return false;
        }
        else {
            crawl = crawl.children.get(ch);
            if(i==n-1 && crawl.end == true){
                return true;
            }

        }
    }
    return false;
}
private void add(String word) {
    Node crawl = root;
    int n = word.length();
    for(int i=0;i<n;i++){
        char ch = word.charAt(i);
        if(crawl.children.containsKey(ch)){
            crawl = crawl.children.get(ch);
        }
        else {
            crawl.children.put(ch, new Node(false));
            Node temp = crawl.children.get(ch);
            if(i == n-1){
                temp.end = true;
            }
            crawl = temp;
            System.out.println(ch + "      " + crawl.end);

        }
    }
}

}

回复于 2024-04-29T09:12:29+08:00

在这里我的实施：

public class Tries {
private static class Leaf {
    private Leaf(char c) {
        this.c=c;
    }
    char c;
    int counter = 1;
    List<Leaf> leaves = new ArrayList<>(10);
}
private Leaf root = new Leaf('0');
public void add(String word) {
    Leaf current = root;
    Leaf newLeaf = null;
    for (char c : word.toCharArray()) {
        boolean found = false;
        for (Leaf leaf : current.leaves) {
            if (leaf.c == c) {
                current = leaf;
                current.counter++;
                found=true;
                break;
            }
        }
        if (!found) {
            newLeaf = new Leaf(c);
            current.leaves.add(newLeaf);
            current = newLeaf;
        }
    }
}
public int find(String partial) {
    Leaf current = root;
    for (char c : partial.toCharArray()) {
        boolean found = false;
        for (Leaf leaf : current.leaves) {
            if (leaf.c == c) {
                current=leaf;
                found=true;
                break;
            }
        }
        if(!found) return 0;
    }
    return current.counter;
}

public boolean hasWord(String partial) {
    return find(partial)>0;
    }
}

回复于 2024-04-29T09:12:29+08:00

1

This Trie Map实现has (called get in this implementation)和toString

回复于 2024-04-29T09:12:29+08:00

这是简单的java实现，不使用任何其他数据结构

import java.util.ArrayList;
import java.util.List;

class Trie {

    private static Node root = new Node(' ', false);

    static int getIndex(char x) {
        return ((int) x) - ((int) 'a');
    }

    static class Node {
        char data;
        boolean isLeaf;
        Node[] children;

        Node(char data, boolean leaf) {
            this.data = data;
            this.isLeaf = leaf;
            this.children = new Node[26];
        }

    }

    static void insert(String data, Node root) {
        if (data == null || data.length() == 0) {
            return;
        }
        Node child = root.children[getIndex(data.charAt(0))];
        if (child == null) {
            Node node = new Node(data.charAt(0), data.length() == 1);
            root.children[getIndex(data.charAt(0))] = node;
            if (data.length() > 1) {
                insert(data.substring(1, data.length()), node);
            }
        } else {
            if (data.length() == 1) {
                child.isLeaf = true;
            } else {
                insert(data.substring(1, data.length()), child);
            }
        }
    }

    static boolean find(String data, Node root) {
        if (data == null || data.length() == 0) {
            return true;
        }
        char x = data.charAt(0);
        //note that first node ie root is just dummy, it just holds important
        Node node = root.children[getIndex(x)];
        if (node == null) {
            return false;
        } else {
            if (data.length() == 1) {
                return node.isLeaf;
            } else {
                return find(data.substring(1, data.length()), node);
            }
        }
    }

    static boolean delete(String data, Node root) {
        if (data == null || data.length() == 0) {
            return false;
        }
        char x = data.charAt(0);
        //note that first node ie root is just dummy, it just holds important
        Node node = root.children[getIndex(x)];
        if (node == null) {
            return false;
        } else {
            if (data.length() == 1) {
                node.isLeaf = false;
                boolean allNull = true;
                for (Node node1 : node.children) {
                    allNull = allNull && node1 == null;
                }
                return allNull;
            } else {
                boolean delete = delete(data.substring(1, data.length()), node);
                if (delete) {
                    node.children[getIndex(x)] = null;
                    if(node.isLeaf){
                        return false;
                    }
                    boolean allNull = true;
                    for (Node node1 : node.children) {
                        allNull = allNull && node1 == null;
                    }
                    return allNull;                }
            }
        }
        return false;
    }


    private static List<String> strings = new ArrayList<>();

    private static List<String> getAll() {
        strings = new ArrayList<String>();
        findAllDFS(root, "");
        return strings;
    }

    private static void findAllDFS(Node node, String old) {
        if (node != null) {
            if (node.data != ' ') {
                old = old + node.data;
            }
            if (node.isLeaf) {
                strings.add(old);
            }
            for (Node node1 : node.children) {
                findAllDFS(node1, old);
            }
        }
    }

    public static void main(String[] args) {
        insert("abc", root);
        insert("xyz", root);
        insert("abcd", root);
        insert("abcde", root);


        delete("abcd", root);

 /*       System.out.println(find("abc", root));
        System.out.println(find("abcd", root));
        System.out.println(find("ab", root));
        System.out.println(find("xyz", root));*/


        System.out.println(getAll());
    }


}

回复于 2024-04-29T09:12:29+08:00

Trie实施

6 回答

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