トライ

トライ データ構造の実装

トライデータ構造の説明

class Node{
    Node [] node = new Node[26];
    boolean flag;
    public Node(){

    }
    public boolean containsKey(char c){
        return node[c-'a']!=null;
    }
    public void put(char c, Node n){
        node[c-'a']  = n;
    }
    public Node get(char c){
        return node[c-'a'];
    }
    public void setFlag() {
        this.flag = true;
    }
    public boolean getFlag(){
        return this.flag;
    }
}

class トライ {
    Node root;
    public トライ() {
        root = new Node();
    }
    //will take tc : O(len) of the word
    public void insert(String word) {
        Node node  = root;
        for(int i =0;i<word.length if node.put node node.setflag take tc : o of the word public boolean search for i="0;i<word.length();i++){" return false node.getflag prefix startswith true your object will be instantiated and called as such: obj="new" obj.insert param_2="obj.search(word);" param_3="obj.startsWith(prefix);">



<h2>
  
  
  データ構造を試す II
</h2>

<p>より理解を深めるためのストリバーの説明<br>
</p>

<pre class="brush:php;toolbar:false">import java.util.* ;
import java.io.*; 

class Node {
    Node node[] = new Node[26];
    int endWith = 0;// will keep track of no. of words ending with this word
    int countPrefix=0;// will keep track of no. of words starting with this word
    public Node(){

    }
    public boolean containsKey(char c){
        return node[c-'a']!=null;
    }
    public void put(char c, Node n){
        node[c-'a'] = n;
    }
    public Node get(char c){
        return node[c-'a'];
    }
    public void incrementCountPrefix() {
        ++this.countPrefix;
    }
    public void decrementCountPrefix(){
        --this.countPrefix;
    }
    public void incrementEndWith(){
        ++this.endWith;
    }
    public void deleteEndWith(){
        --this.endWith;
    }
    public int getCountPrefix(){
        return this.countPrefix;
    }
    public int getEndWith(){
        return this.endWith;
    }


}
public class トライ {
    Node root;
    public トライ() {
        // Write your code here.
        root = new Node();
    }

    public void insert(String word) {
        Node node = root;
        for(int i =0;i<word.length if node.put node node.incrementcountprefix node.incrementendwith public int countwordsequalto word write your code here. for i="0;i<word.length();i++){" else return node.getendwith will tell how many strings are with given countwordsstartingwith node.getcountprefix it starting void erase node.decrementcountprefix node.deleteendwith>



<h2>
  
  
  完全な文字列
</h2>



<pre class="brush:php;toolbar:false">// tc : O(n*l)

import java.util.* ;
import java.io.*; 

class Node{
    Node[] node = new Node[26];
    boolean flag;
    public Node(){}
    public void put(char c , Node n){
        node[c-'a'] = n;
    }
    public boolean containsKey(char c){
        return node[c-'a']!=null;
    }
    public Node get(char c){
        return node[c-'a'];
    }
    public void setEnd(){
        this.flag = true;
    }
    public boolean isEnd(){
        return this.flag;
    }
}

class トライ{
    Node root;
    public トライ(){
        root = new Node();
    }
    public boolean checkIfPrefixPresent(String s){
      Node node = root;
      boolean flag= true;
      for(int i = 0;i<s.length char c="s.charAt(i);" if return false node="node.get(c);" flag="flag" node.isend this will check the substring is also a string from list of strings line work here because any not present as in trie then s won be complete and we can only public void insert for i="0;i<s.length();i++){" node.put new node.setend setting end current class solution static root completestring n all data structure : trie.insert out comeplete among s.length selection a.compareto b lexographically else s.compareto completestring.equals>



<h2>
  
  
  個別の部分文字列を数える
</h2>

<p>Tc: <br> に異なる一意の部分文字列を挿入するための O(n^2)
データ構造を試す<br>
</p>

<pre class="brush:php;toolbar:false">
import java.util.ArrayList;

public class Solution 
{
    Node root;
    static int count;
    public Solution(){
        root = new Node();
    }

    public static int countDistinctSubstrings(String s) 
    {
        count = 0;
        //    Write your code here.
        Solution sol = new Solution();
        for(int i =0;i




          

            
  

            
        

以上がトライの詳細内容です。詳細については、PHP 中国語 Web サイトの他の関連記事を参照してください。