【DFS + Backtracking】LeetCode 212. Word Search II

LeetCode 212. Word Search II

Solution1:我的答案
暴力搜索。。基于第79题的答案，真的是非常之慢啊！！！
快的方法均是基于字典树的方法，真是复杂。。

class Solution {
public:
    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        set<string> res;
        if (!board.size() || !board[0].size() || !words.size())
            return {};
        int m = board.size(), n = board[0].size();
        for (auto word: words) {
            bool symbol = false;
            for (int i = 0; i < m; i++) {
                for (int j = 0; j < n; j++) {
                    vector<vector<int> > visited(m, vector<int>(n, 0));//0:未用;1:用过
                    string temp;
                    int len = 0;
                    symbol = my_search(board, i, j, len, word, visited);
                    if (symbol) res.insert(word);
                }
            }
        }
        return vector<string> (res.begin(), res.end());
    }

    bool my_search(vector<vector<char>>& board, int i, int j, int len,
                   string& word, vector<vector<int> >& visited) {
        if (len == word.size()) 
            return true;
        if (i >= 0 && i < board.size() && j >= 0 
            && j < board[0].size() && visited[i][j] == 0 
            && board[i][j] == word[len]) {
            len++;
            visited[i][j] = 1;
            if (my_search(board, i - 1, j, len, word, visited) 
                || my_search(board, i + 1, j, len, word, visited)
                || my_search(board, i, j - 1, len, word, visited)
                || my_search(board, i, j + 1, len, word, visited))
                return true;
            len--;
            visited[i][j] = 0;
        }
        return false;
    }
};