212. Word Search II

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.

struct Trie{
    Trie *next[26];
    int index;
    Trie() : index(-1){
        memset(next,0,sizeof(next));
    }

    void insert(const char *key, int idx){
        if(eof(*key)!=0){
            if(next[*key - 'a']==0)
                next[*key - 'a'] = new Trie();
            next[*key - 'a']->insert(key+1, idx);
        }
        else
            index = idx;
    }

    bool eof(char c){
        return ('a'<=c&&c<='z');
    }
};
class Solution {
    unordered_set<string> w;
    bool visit[12][12]{false,};
    int dy[4]{-1,0,1,0}, dx[4]{0,1,0,-1};
    void dfs(vector<vector<char>>& b, int &n, int &m, vector<string>& w, vector<string>& res, Trie* trie, int y, int x) {
        auto key = b[y][x];
        if(b[y][x] == '#' || !(trie->next[key - 'a'])) return;
        Trie* nxt = trie = trie->next[key - 'a'];
        b[y][x] = '#';
        if(nxt->index != -1) {
            res.push_back(w[nxt->index]);
            nxt->index = -1;
        }
        for(int i = 0; i < 4; i++) {
            int ny = y + dy[i], nx = x + dx[i];
            if(0 <= nx and nx < m and 0 <= ny and ny < n) {
                dfs(b,n,m,w,res,nxt,ny,nx);
            }
        }
        b[y][x] = key;
    }

public:
    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        int n = board.size(), m = board[0].size();
        Trie* trie = new Trie();
        for(int i = 0; i < words.size(); i++) {
            trie->insert(words[i].c_str(), i);
        }
        vector<string> res;
        for(int i = 0; i < n; i++)
            for(int j = 0; j < m; j++) {
                dfs(board,n,m,words,res,trie,i,j);
            }

        return res;
    }
};