Power button (LeetCode) 212. The word search II (2022.07.31)

给定一个 m x n 二维字符网格 board 和一个单词（字符串）列表 words, 返回所有二维网格上的单词 .

单词必须按照字母顺序,通过 相邻的单元格 内的字母构成,其中“相邻”单元格是那些水平相邻或垂直相邻的单元格.同一个单元格内的字母在一个单词中不允许被重复使用.

示例 1：

输入：board = [[“o”,“a”,“a”,“n”],[“e”,“t”,“a”,“e”],[“i”,“h”,“k”,“r”],[“i”,“f”,“l”,“v”]], words = [“oath”,“pea”,“eat”,“rain”]
输出：[“eat”,“oath”]
示例 2：

输入：board = [[“a”,“b”],[“c”,“d”]], words = [“abcb”]
输出：[]

提示：

m == board.length
n == board[i].length
1 <= m, n <= 12
board[i][j] 是一个小写英文字母
1 <= words.length <= 3 * 104
1 <= words[i].length <= 10
words[i] 由小写英文字母组成
words 中的所有字符串互不相同

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/word-search-ii

方法一：回溯+字典树

C++提交内容：

struct TrieNode {
    
    string word;
    unordered_map<char,TrieNode *> children;
    TrieNode() {
    
        this->word = "";
    }   
};

void insertTrie(TrieNode * root,const string & word) {
    
    TrieNode * node = root;
    for (auto c : word){
    
        if (!node->children.count(c)) {
    
            node->children[c] = new TrieNode();
        }
        node = node->children[c];
    }
    node->word = word;
}

class Solution {
    
public:
    int dirs[4][2] = {
    {
    1, 0}, {
    -1, 0}, {
    0, 1}, {
    0, -1}};

    bool dfs(vector<vector<char>>& board, int x, int y, TrieNode * root, set<string> & res) {
    
        char ch = board[x][y];        
        if (!root->children.count(ch)) {
    
            return false;
        }
        root = root->children[ch];
        if (root->word.size() > 0) {
    
            res.insert(root->word);
        }

        board[x][y] = '#';
        for (int i = 0; i < 4; ++i) {
    
            int nx = x + dirs[i][0];
            int ny = y + dirs[i][1];
            if (nx >= 0 && nx < board.size() && ny >= 0 && ny < board[0].size()) {
    
                if (board[nx][ny] != '#') {
    
                    dfs(board, nx, ny, root,res);
                }
            }
        }
        board[x][y] = ch;

        return true;      
    }

    vector<string> findWords(vector<vector<char>> & board, vector<string> & words) {
    
        TrieNode * root = new TrieNode();
        set<string> res;
        vector<string> ans;

        for (auto & word: words){
    
            insertTrie(root,word);
        }
        for (int i = 0; i < board.size(); ++i) {
    
            for (int j = 0; j < board[0].size(); ++j) {
    
                dfs(board, i, j, root, res);
            }
        }        
        for (auto & word: res) {
    
            ans.emplace_back(word);
        }
        return ans;        
    }
};