212. 单词搜索 II (Trie&dfs)

Posted 2021-10-06 Harris-H

212. 单词搜索 II (Trie&dfs)

对方格每个位置进行$dfs$，同时用一个$Trie$当前走的字符串是否为$Trie$的前缀。

为了更快，当我们找到一个字符串时，就把它从$Trie$删去。

struct TrieNode {
    string word;
    int num;
    unordered_map<char, TrieNode *> children;
    TrieNode() {
        this->word = "";
        num=0;
    }   
};

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

    }
}
class Solution {
public:
    int dirs[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    TrieNode * _root = new TrieNode();
    bool dfs(vector<vector<char>>& board, int x, int y, TrieNode * root, vector<string> & res) {
        if(!root->num) return false; //这个剪枝很重要
        char ch = board[x][y];        
        if (!root->children.count(ch)) {
            return false;
        }
        root = root->children[ch];
        if (root->word.size() > 0 ) {
            res.push_back(root->word);
            erase(_root,root->word);
            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) {
        vector<string> res;
        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);   
        return res;        
    }
};

时间复杂度：$O(nm{3}^{l-1})$，$l$是最长字符串长度，因为同一个单词的一个格子不能多次使用，所以每次最多3个方向选择。