java_implement_trie

A trie (pronounced as "try") or prefix tree is a tree data structure used to efficiently store and retrieve keys in a dataset of strings. There are various applications of this data structure, such as autocomplete and spellchecker.

Implement the Trie class:

• Trie() Initializes the trie object.
• void insert(String word) Inserts the string word into the trie.
• boolean search(String word) Returns true if the string word is in the trie (i.e., was inserted before), and false otherwise.
• boolean startsWith(String prefix) Returns true if there is a previously inserted string word that has the prefix prefix, and false otherwise.

Examples

Example 1:

Input
["Trie", "insert", "search", "search", "startsWith", "insert", "search"]
[[], ["apple"], ["apple"], ["app"], ["app"], ["app"], ["app"]]
Output
[null, null, true, false, true, null, true]

Explanation
Trie trie = new Trie();
trie.insert("apple");
trie.search("apple");   // return True
trie.search("app");     // return False
trie.startsWith("app"); // return True
trie.insert("app");
trie.search("app");     // return True

Constraints:

• 1 <= word.length, prefix.length <= 2000
• word and prefix consist only of lowercase English letters.
• At most $3*10^4$ calls in total will be made to insert, search, and startsWith.

解析

題目要我們實作一個可以儲存所有小寫英文字的一個 Trie結構

需要實作出以下 method:

1. Constructor: 用來來建構 Trie
2. insert(word string): 用來儲存 word
3. search(word string) bool: 用來察看是否有儲存過這個 word
4. startsWith(prefix string) bool: 用來察看是否有儲存過包含 prefix 的 word

首先知道 Trie 是一種特殊的 Tree

目前需要儲存所有字元是小寫英文 a-z 組成的字串

為了節省空間所以會是已字元為單位來做存儲

另外是可以透過把 字元當成每個 node 的 edge 上 key

如下圖：

每個結點都可以透過 key 來找到下一個結點

這個結構可以使用 hashmap 來實作

而沒每個結點除了這個用來紀錄 child 對應的 hashmap 外

需要一個布林值來紀錄這個結點是否為最後一個結點

如上圖的 Trie 雖然有 appl 的 prefix 但是卻沒有 appl 這個字

程式碼

public class Trie {
  static class TrieNode {
    boolean EndOfWord = false;
    HashMap<Character, TrieNode> Children = new HashMap<>();
  }
  TrieNode root;
  public Trie() {
    root = new TrieNode();
  }
  public void insert(String word) {
    TrieNode cur = root;
    int size = word.length();
    for (int pos = 0; pos < size; pos++) {
      char ch = word.charAt(pos);
      if (!cur.Children.containsKey(ch)) {
        cur.Children.put(ch, new TrieNode());
      }
      cur = cur.Children.get(ch);
    }
    cur.EndOfWord = true;
  }
  public boolean search(String word) {
    TrieNode cur = root;
    int size = word.length();
    for (int pos = 0; pos < size; pos++) {
      char ch = word.charAt(pos);
      if (!cur.Children.containsKey(ch)) {
        return false;
      }
      cur = cur.Children.get(ch);
    }
    return cur.EndOfWord;
  }
  public boolean startsWith(String prefix) {
    TrieNode cur = root;
    int size = prefix.length();
    for (int pos = 0; pos < size; pos++) {
      char ch = prefix.charAt(pos);
      if (!cur.Children.containsKey(ch)) {
        return false;
      }
      cur = cur.Children.get(ch);
    }
    return true;
  }
}

困難點

1. 轉換紀錄 edge的值當作 index

Solve Point

• Understand What problem need to solve
• Analysis complexity