212. Word Search II

Hard (Khó) C++ 🔗 Xem trên LeetCode

📋 Đề Bài

Given an m x n board of characters and a list of strings words, return all words on the board.

Each word must be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.

 

Example 1:

Input: board = [["o","a","a","n"],["e","t","a","e"],["i","h","k","r"],["i","f","l","v"]], words = ["oath","pea","eat","rain"]
Output: ["eat","oath"]

Example 2:

Input: board = [["a","b"],["c","d"]], words = ["abcb"]
Output: []

 

Constraints:

  • m == board.length
  • n == board[i].length
  • 1 <= m, n <= 12
  • board[i][j] is a lowercase English letter.
  • 1 <= words.length <= 3 * 104
  • 1 <= words[i].length <= 10
  • words[i] consists of lowercase English letters.
  • All the strings of words are unique.

🧠 Thuật Toán & Kỹ Thuật

DFS (Tìm kiếm theo chiều sâu)Bit Manipulation (Thao tác bit)Trie (Cây tiền tố)String (Chuỗi)
⏱️ Thời gian O(V+E)
💾 Không gian O(V)

💻 Lời Giải

C++ 0212-word-search-ii.cpp 
struct TrieNode {
    TrieNode* children[26] = {};
    string* word;
    void addWord(string& word) {
        TrieNode* cur = this;
        for (char c : word) {
            c -= 'a';
            if (cur->children[c] == nullptr) cur->children[c] = new TrieNode();
            cur = cur->children[c];
        }
        cur->word = &word;
    }
};

class Solution {
public:
    int m, n;
    int DIR[5] = {0, 1, 0, -1, 0};
    vector<string> ans;
    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        m = board.size(); n = board[0].size();
        TrieNode trieNode;
        for (string& word : words) trieNode.addWord(word);
        
        for (int r = 0; r < m; ++r)
            for (int c = 0; c < n; ++c)
                dfs(board, r, c, &trieNode);
        return ans;
    }
    void dfs(vector<vector<char>>& board, int r, int c, TrieNode* cur) {
        if (r < 0 || r == m || c < 0 || c == n || board[r][c] == '#' || cur->children[board[r][c]-'a'] == nullptr) return;
        char orgChar = board[r][c];
        cur = cur->children[orgChar - 'a'];
        if (cur->word != nullptr) {
            ans.push_back(*cur->word);
            cur->word = nullptr;
        }
        board[r][c] = '#';
        for (int i = 0; i < 4; ++i) dfs(board, r + DIR[i], c + DIR[i+1], cur);
        board[r][c] = orgChar;
    }
};