212. 单词搜索 II（字典树/前缀树）

给定一个 m x n 二维字符网格 board 和一个单词（字符串）列表 words， 返回所有二维网格上的单词 。

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 1：

输入：board = [["o","a","a","n"],["e","t","a","e"],["i","h","k","r"],["i","f","l","v"]], words = ["oath","pea","eat","rain"]
输出：["eat","oath"]

示例 2：

输入：board = [["a","b"],["c","d"]], words = ["abcb"]
输出：[]

提示：

• m == board.length
• n == board[i].length
• 1 <= m, n <= 12
• board[i][j] 是一个小写英文字母
• 1 <= words.length <= 3 * 104
• 1 <= words[i].length <= 10
• words[i] 由小写英文字母组成
• words 中的所有字符串互不相同

字典树/前缀树，模板题。最初版本见题号208，实现字典树。进阶如题号472，字典树+dfs记忆化搜索这类。

class Solution {
public:
struct Trie
{
    Trie* child[26];
    string word = "";
    Trie() {
        for (int i = 0; i < 26; i++)
            child[i] = nullptr;
    }
};
    vector<string> ans;
    
    void dfs(vector<vector<char>>& board,Trie* t,int i,int j)
    {
        char c=board[i][j];
        if(c=='*'||t->child[c-'a']==nullptr) return;
        t=t->child[c-'a'];
        if(t->word!="")
        {
            ans.push_back(t->word);
            t->word="";
        }
        board[i][j]='*';
        if(i+1<board.size())dfs(board,t,i+1,j);
        if(i-1>=0)dfs(board,t,i-1,j);
        if(j+1<board[i].size())dfs(board,t,i,j+1);
        if(j-1>=0)dfs(board,t,i,j-1);
        board[i][j]=c;
        return;
    }
    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        Trie* t=new Trie();
        for(int i=0;i<words.size();i++)
        {
            Trie* cur=t;
            for(int j=0;j<words[i].size();j++)
            {
                if(cur->child[words[i][j]-'a']==nullptr)
                {
                    cur->child[words[i][j]-'a']=new Trie();
                }
                cur=cur->child[words[i][j]-'a'];
            }
            cur->word=words[i];
        }
       
        for(int i=0;i<board.size();i++)
        {
            for(int j=0;j<board[i].size();j++)
            {
                dfs(board,t,i,j);
            }
        }
        return ans;
    }
};