LeetCode //C - 211. Design Add and Search Words Data Structure

Design a data structure that supports adding new words and finding if a string matches any previously added string.

Implement the WordDictionary class:

  • WordDictionary() Initializes the object.
  • void addWord(word) Adds word to the data structure, it can be matched later.
  • bool search(word) Returns true if there is any string in the data structure that matches word or false otherwise. word may contain dots '.' where dots can be matched with any letter.
Example:

Input:

"WordDictionary","addWord","addWord","addWord","search","search","search","search"

\[\],\["bad"\],\["dad"\],\["mad"\],\["pad"\],\["bad"\],\[".ad"\],\["b..."\]

Output:

null,null,null,null,false,true,true,true

Explanation:

WordDictionary wordDictionary = new WordDictionary();

wordDictionary.addWord("bad");

wordDictionary.addWord("dad");

wordDictionary.addWord("mad");

wordDictionary.search("pad"); // return False

wordDictionary.search("bad"); // return True

wordDictionary.search(".ad"); // return True

wordDictionary.search("b..."); // return True

Constraints:
  • 1 <= word.length <= 25
  • word in addWord consists of lowercase English letters.
  • word in search consist of '.' or lowercase English letters.
  • There will be at most 2 dots in word for search queries.
  • At most 1 0 4 10^4 104 calls will be made to addWord and search.

From: LeetCode

Link: 211. Design Add and Search Words Data Structure


Solution:

Ideas:

1. TrieNode Structure:

Each node in the Trie is represented by a TrieNode structure. It has the following components:

  • An array of pointers, children, where each pointer corresponds to a letter in the English alphabet (26 lowercase letters).
  • A boolean flag, isEndOfWord, to signify whether a word ends at this node.

2. WordDictionary Structure:

The WordDictionary itself is represented by a structure, which holds a pointer to the root node of the Trie.

3. wordDictionaryCreate:

This function initializes the WordDictionary object and allocates memory for the root node of the Trie.

4. wordDictionaryAddWord:

This function is used to insert words into the Trie. For each character in the word, it traverses down the Trie, creating new nodes if needed, until the end of the word is reached, at which point it sets the isEndOfWord flag to true.

5. wordDictionarySearch and searchHelper:

  • The wordDictionarySearch function is used to search for a word in the Trie, with support for the . character, which can match any letter.
  • It calls a helper function searchHelper, which performs a recursive search to handle the . character.
  • If the searchHelper encounters a . character, it recursively checks all its children.
  • If it can traverse the entire word and reach a node where isEndOfWord is true, it returns true; otherwise, it returns false.

6. wordDictionaryFree and freeNode:

  • These functions deallocate the memory used by the WordDictionary and its nodes.
  • freeNode is a recursive function that frees all the child nodes before freeing the parent node.
Code:
c 复制代码
#define ALPHABET_SIZE 26

typedef struct TrieNode {
    struct TrieNode *children[ALPHABET_SIZE];
    bool isEndOfWord;
} TrieNode;

typedef struct {
    TrieNode *root;
} WordDictionary;

TrieNode* createNode() {
    TrieNode *newNode = (TrieNode *)calloc(1, sizeof(TrieNode));
    return newNode;
}

WordDictionary* wordDictionaryCreate() {
    WordDictionary *dict = (WordDictionary *)malloc(sizeof(WordDictionary));
    dict->root = createNode();
    return dict;
}

void wordDictionaryAddWord(WordDictionary* obj, char * word) {
    TrieNode *node = obj->root;
    for (int i = 0; word[i] != '\0'; i++) {
        int index = word[i] - 'a';
        if (!node->children[index])
            node->children[index] = createNode();
        node = node->children[index];
    }
    node->isEndOfWord = true;
}

bool searchHelper(TrieNode *node, char *word) {
    for (int i = 0; word[i] != '\0'; i++) {
        if (word[i] == '.') {
            for (int j = 0; j < ALPHABET_SIZE; j++) {
                if (node->children[j] && searchHelper(node->children[j], word + i + 1))
                    return true;
            }
            return false;
        } else {
            int index = word[i] - 'a';
            if (!node->children[index])
                return false;
            node = node->children[index];
        }
    }
    return node->isEndOfWord;
}

bool wordDictionarySearch(WordDictionary* obj, char * word) {
    return searchHelper(obj->root, word);
}

void freeNode(TrieNode *node) {
    for(int i = 0; i < ALPHABET_SIZE; i++)
        if(node->children[i])
            freeNode(node->children[i]);
    free(node);
}

void wordDictionaryFree(WordDictionary* obj) {
    if(!obj) return;
    freeNode(obj->root);
    free(obj);
}

/**
 * Your WordDictionary struct will be instantiated and called as such:
 * WordDictionary* obj = wordDictionaryCreate();
 * wordDictionaryAddWord(obj, word);
 
 * bool param_2 = wordDictionarySearch(obj, word);
 
 * wordDictionaryFree(obj);
*/
相关推荐
To_OC13 小时前
LC 1 两数之和:面试第一道必考题,暴力解法直接被面试官 pass
javascript·算法·leetcode
鱼鱼不愚与18 小时前
《原来如此 | 第01期:为什么导航软件能预测红绿灯倒计时?》
算法
复杂网络1 天前
论最小 Agent 计算机的形态
算法
kisshyshy2 天前
🍦 雪糕、食堂、火车厢:三幅漫画吃透栈、队列与链表
javascript·算法
猿人谷2 天前
不只是 CPU 阈值:STAR 如何用 GAT + Transformer 做容器级自动扩缩容?
人工智能·算法
复杂网络2 天前
Stable Diffusion 视觉大模型微调技术深度调研
算法
复杂网络2 天前
基于 Stable Diffusion 架构的视觉大模型代表性工作与原理深度解析
算法
MrZhao4002 天前
Agent Loop 如何用 Hook 扩展:权限、日志与工具拦截
算法
MrZhao4002 天前
Agent 为什么需要 Skills:别把所有知识都塞进 system prompt
算法