📜  使用 Trie 进行拼写检查

📅  最后修改于: 2021-09-02 06:01:13             🧑  作者: Mango

给定一个字符串数组str[]和一个字符串key ,任务是检查的拼写是否正确。如果发现是真的,则打印“YES” 。否则,打印建议的正确拼写。

例子:

方法:使用Trie可以解决这个问题。我们的想法是遍历字符串,STR [阵列],然后插入字符串入Trie树使得Trie树中的每个节点包含字符串和一个布尔值的字符,以检查是否该字符是字符串的最后一个字符或不是。请按照以下步骤解决问题:

  • 初始化一个Trie树,比方说,使得Trie树中的每个节点都包括一个字符串的字符和一个布尔值的检查,如果字符是字符串或不是最后一个字符。
  • 遍历字符串数组arr[] ,并将所有字符串插入到 Trie 中。
  • 最后,遍历字符串key 。对于每个i字符,检查该字符是否存在于Trie 中。如果发现为真,则移动到 Trie 的下一个节点。
  • 否则,打印所有可能的字符串,其前缀是字符串key

下面是上述方法的实现:

C++
// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Structure of a Trie node
struct TrieNode {
 
    // Store address of a character
    TrieNode* Trie[256];
 
    // Check if the character is
    // last character of a string or not
    bool isEnd;
 
    // Constructor function
    TrieNode()
    {
 
        for (int i = 0; i < 256; i++) {
 
            Trie[i] = NULL;
        }
        isEnd = false;
    }
};
 
// Function to insert a string into Trie
void InsertTrie(TrieNode* root, string s)
{
 
    TrieNode* temp = root;
 
    // Traverse the string, s
    for (int i = 0; i < s.length(); i++) {
 
        if (temp->Trie[s[i]] == NULL) {
 
            // Initialize a node
            temp->Trie[s[i]] = new TrieNode();
        }
 
        // Update temp
        temp = temp->Trie[s[i]];
    }
 
    // Mark the last character of
    // the string to true
    temp->isEnd = true;
}
 
// Function to print suggestions of the string
void printSuggestions(TrieNode* root, string res)
{
 
    // If current character is
    // the last character of a string
    if (root->isEnd == true) {
 
        cout << res << " ";
    }
 
    // Iterate over all possible
    // characters of the string
    for (int i = 0; i < 256; i++) {
 
        // If current character
        // present in the Trie
        if (root->Trie[i] != NULL) {
 
            // Insert current character
            // into Trie
            res.push_back(i);
            printSuggestions(root->Trie[i], res);
            res.pop_back();
        }
    }
}
 
// Function to check if the string
// is present in Trie or not
bool checkPresent(TrieNode* root, string key)
{
 
    // Traverse the string
    for (int i = 0; i < key.length(); i++) {
 
        // If current character not
        // present in the Trie
        if (root->Trie[key[i]] == NULL) {
 
            printSuggestions(root, key.substr(0, i));
 
            return false;
        }
 
        // Update root
        root = root->Trie[key[i]];
    }
    if (root->isEnd == true) {
 
        return true;
    }
    printSuggestions(root, key);
 
    return false;
}
 
// Driver Code
int main()
{
 
    // Given array of strings
    vector str = { "gee", "geeks", "ape",
                           "apple", "geeksforgeeks" };
 
    string key = "geek";
 
    // Initialize a Trie
    TrieNode* root = new TrieNode();
 
    // Insert strings to trie
    for (int i = 0; i < str.size(); i++) {
        InsertTrie(root, str[i]);
    }
 
    if (checkPresent(root, key)) {
 
        cout << "YES";
    }
    return 0;
}


Java
// Java program to implement
// the above approach
import java.io.*;
 
// Structure of a Trie node
class TrieNode
{
     
    // Store address of a character
    TrieNode Trie[];
 
    // Check if the character is
    // last character of a string or not
    boolean isEnd;
 
    // Constructor function
    public TrieNode()
    {
        Trie = new TrieNode[256];
        for(int i = 0; i < 256; i++)
        {
            Trie[i] = null;
        }
        isEnd = false;
    }
}
 
class GFG{
 
// Function to insert a string into Trie
static void InsertTrie(TrieNode root, String s)
{
    TrieNode temp = root;
 
    // Traverse the string, s
    for(int i = 0; i < s.length(); i++)
    {
        if (temp.Trie[s.charAt(i)] == null)
        {
             
            // Initialize a node
            temp.Trie[s.charAt(i)] = new TrieNode();
        }
 
        // Update temp
        temp = temp.Trie[s.charAt(i)];
    }
 
    // Mark the last character of
    // the string to true
    temp.isEnd = true;
}
 
// Function to print suggestions of the string
static void printSuggestions(TrieNode root, String res)
{
 
    // If current character is
    // the last character of a string
    if (root.isEnd == true)
    {
        System.out.print(res + " ");
    }
 
    // Iterate over all possible
    // characters of the string
    for(int i = 0; i < 256; i++)
    {
         
        // If current character
        // present in the Trie
        if (root.Trie[i] != null)
        {
             
            // Insert current character
            // into Trie
            res += (char)i;
            printSuggestions(root.Trie[i], res);
            res = res.substring(0, res.length() - 2);
        }
    }
}
 
// Function to check if the string
// is present in Trie or not
static boolean checkPresent(TrieNode root, String key)
{
 
    // Traverse the string
    for(int i = 0; i < key.length(); i++)
    {
         
        // If current character not
        // present in the Trie
        if (root.Trie[key.charAt(i)] == null)
        {
            printSuggestions(root, key.substring(0, i));
            return false;
        }
 
        // Update root
        root = root.Trie[key.charAt(i)];
    }
    if (root.isEnd == true)
    {
        return true;
    }
    printSuggestions(root, key);
 
    return false;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array of strings
    String str[] = { "gee", "geeks", "ape", "apple",
                     "geeksforgeeks" };
 
    String key = "geek";
 
    // Initialize a Trie
    TrieNode root = new TrieNode();
 
    // Insert strings to trie
    for(int i = 0; i < str.length; i++)
    {
        InsertTrie(root, str[i]);
    }
 
    if (checkPresent(root, key))
    {
        System.out.println("YES");
    }
}
}
 
// This code is contributed by Dharanendra L V.


输出:
geeks geeksforgeeks

时间复杂度: O(N * M),其中 M 是字符串的最大长度
辅助空间: O(N * 256)

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