给出一个2D板和字典中的单词列表，找到棋盘上的所有单词。每个单词必须由顺序相邻单元格的字母构成。不能重复使用

本题源自leetcode   212

-----------------------------------------------------------------------

思路：

构造一个Trie单词查找树。然后用递归遍历棋盘。找到所有的单词。

代码：

   class Trie{    public:        int idx;        bool leaf;        Trie* children[26];        Trie(){            this->idx = 0;  //如果当前节点是结尾，idx 是这个字符串在字典中的下标            this->leaf = false;   //表示是否是一个字符串的结尾            fill_n(this->children,26,nullptr);  //指向其从a-z的开始子串的指针。        }        };public:    void insertWords(Trie* root,vector<string>& words,int idx){        int len = words[idx].size();        int i = 0;        while(i < len){            if(root->children[words[idx][i] - 'a'] == NULL)                root->children[words[idx][i] - 'a'] = new Trie();            root = root->children[words[idx][i] - 'a'];            i++;        }        root->leaf = true;        root->idx = idx;    }    Trie* buildTrie(vector<string>& words){        Trie* root = new Trie();        for(int i = 0; i < words.size(); i++){            insertWords(root,words,i);        }        return root;    }    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {        vector<string> res;        int row = board.size();        if(row == 0)            return res;        int col = board[0].size();        if(col == 0)            return res;        int n = words.size();        if(n == 0)            return res;                Trie* root = buildTrie(words);        for(int i = 0; i < row; i++){            for(int j = 0; j < col && n > res.size(); j++){                dfs(board,words,res,root,i,j,row,col);            }        }        return res;    }    void dfs(vector<vector<char>>& board,vector<string>& words,vector<string>& res,Trie* root,int i,int j,int row,int col){        if(board[i][j] == '#')  //已经访问过            return;        char tmp = board[i][j];        if(root->children[tmp - 'a'] == NULL)            return ;        if(root->children[tmp - 'a']->leaf){            res.push_back(words[root->children[tmp - 'a']->idx]);            root->children[tmp - 'a']->leaf = false;        }        board[i][j] = '#';        if(i > 0)            dfs(board,words,res,root->children[tmp - 'a'],i-1,j,row,col);        if(i < row - 1)            dfs(board,words,res,root->children[tmp - 'a'],i+1,j,row,col);        if(j > 0)            dfs(board,words,res,root->children[tmp - 'a'],i,j-1,row,col);        if(j < col - 1)            dfs(board,words,res,root->children[tmp - 'a'],i,j+1,row,col);        board[i][j] = tmp;            }