特里树

实现 特里树 数据结构

特里树数据结构的Strider讲解

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 特里树 {
    Node root;
    public 特里树() {
        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>
  
  
  特里树数据结构二
</h2>

<p>奋斗者的解释，以便更好地理解<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 特里树 {
    Node root;
    public 特里树() {
        // 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>
  
  
  完整字符串
</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 特里树{
    Node root;
    public 特里树(){
        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>
  
  
  计算不同子串的个数
</h2>

<p>Tc：O(n^2)，用于在 <br> 中插入不同的唯一子字符串
特里树数据结构<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




          

            
  

            
        

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