How to use a list of rows in a Trie data structure?

Question

I am working on some performance tests in various data structures. My list has a HashMap and Trie data structure. I ended up with HashMap but not sure how to use Trie to solve the problem below -

I have a text file that contains 2 million English words with their frequencies in this format -

hello 100
world 5000
good 2000
bad 9000
...

Now I read this file line by line and save it to HashMap. The first split line goes like a key in HashMap, and the next split line goes like a value in HashMap, and so I can measure the insert with the code below.

Map<String, String> wordTest = new HashMap<String, String>();

try {
        fis = new FileInputStream(FILE_LOCATION);
        reader = new BufferedReader(new InputStreamReader(fis));

        String line = reader.readLine();
        while (line != null) {
        String[] splitString = line.split("\\s+");
        // now put it in HashMap as key value  pair
        wordTest.put(splitString[0].toLowerCase().trim(), splitString[1].trim());

        line = reader.readLine();
    }
}

, Trie, Trie, HashMap? String? , Trie .

: -

TrieImpl

public class TrieImpl {

    //root node
    private TrieNode r;

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

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

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

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

    public static void main(String[] args) {

        TrieImpl t = new TrieImpl();

        System.out.println("Testing some strings");
        t.insert("HELLO"); // how do I pass string and its count
        t.insert("WORLD"); // how do I pass string and its count

    }
}

TrieNode class -

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 "";
    }
}

, , , String?

Answer 1

frequency TrieNode.

public class TrieNode {

    // make child nodes
    private TrieNode[] c;
    // flag for end of word
    private boolean flag = false;
    //stores frequency if flag is set
    private int frequency;

, flag..change

protected void insert(String word, int frequency) {
    int val = word.charAt(0) - 64;
    ..........
    ..........
    // if the value of the child node at val is null, make a new nod
    if (word.length() > 1) {
        c[val].insert(word.substring(1),frequency);
    } else {
        c[val].flag = true;
        c[val].frequency = frequency;
    }
}

. has, , , , , , .

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

------------------------------- EDIT --------------- ---------

has, , getFreq

    public int getFreq(String word) {
        if(has(word))
            return getFreqHelper(word);
        else
            return -1; //this indicates word is not present
    }

    private int getFreqHelper(String word) {
        int val = word.charAt(0) - 64;
        if (word.length() > 1) {
            return c[val].getFreq(word.substring(1));
        } else if (c[val].flag == true && word.length() == 1) {
            return c[val].frequency;
        } else
        return -1;
}

Answer 2

: FrequencyString :

class FrequencyString {
    private String string;
    private int frequency;

    public FrequencyString(String str, int freq) {
        this.string = str;
        this.frequency = freq;
    }

    public getString() {
        return string;
    }

    public getFrequency() {
        return frequency;
    }
}

Trie, FrequencyString. :

TrieImpl:

boolean has(String word);
void insert(String word, int freq);

TrieNode:

boolean has(String word);
void insert(FrequencyString word);

, , has :

Integer find(String word);

find null, , new Integer(result.getFrequency()); ( result - FrequencyString), .