188.单词搜索 II

给定一个二维网格 board 和一个字典中的单词列表 words，找出所有同时在二维网格和字典中出现的单词。
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:

输入: 
words = ["oath","pea","eat","rain"] and board =
[
  ['o','a','a','n'],
  ['e','t','a','e'],
  ['i','h','k','r'],
  ['i','f','l','v']
]
输出: ["eat","oath"]
说明:
你可以假设所有输入都由小写字母 a-z 组成。

代码

class Solution {
public:
    struct TrieNode {
        TrieNode *child[26];
        string str;
        TrieNode() : str("") {
            for (auto &a : child) a = NULL;
        }
    };
    struct Trie {
        TrieNode *root;
        Trie() : root(new TrieNode()) {}
        void insert(string s) {
            TrieNode *p = root;
            for (auto &a : s) {
                int i = a - 'a';
                if (!p->child[i]) p->child[i] = new TrieNode();
                p = p->child[i];
            }
            p->str = s;
        }
    };
    vector findWords(vector >& board, vector& words) {
        vector res;
        if (words.empty() || board.empty() || board[0].empty()) return res;
        vector > visit(board.size(), vector(board[0].size(), false));
        Trie T;
        for (auto &a : words) T.insert(a);
        for (int i = 0; i < board.size(); ++i) {
            for (int j = 0; j < board[i].size(); ++j) {
                if (T.root->child[board[i][j] - 'a']) {
                    search(board, T.root->child[board[i][j] - 'a'], i, j, visit, res);
                }
            }
        }
        return res;
    }
    void search(vector > &board, TrieNode *p, int i, int j, vector > &visit, vector &res) { 
        if (!p->str.empty()) {
            res.push_back(p->str);
            p->str.clear();
        }
        int d[][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        visit[i][j] = true;
        for (auto &a : d) {
            int nx = a[0] + i, ny = a[1] + j;
            if (nx >= 0 && nx < board.size() && ny >= 0 && ny < board[0].size() && !visit[nx][ny] && p->child[board[nx][ny] - 'a']) {
                search(board, p->child[board[nx][ny] - 'a'], nx, ny, visit, res);
            }
        }
        visit[i][j] = false;
    }
};