212. Word Search II

Given a 2D board and a list of words from the dictionary, find all words in the board.

Each word must be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.

Example:

Input: 
words = ["oath","pea","eat","rain"] and board =
[
  ['o','a','a','n'],
  ['e','t','a','e'],
  ['i','h','k','r'],
  ['i','f','l','v']
]

Output: ["eat","oath"]

Note: You may assume that all inputs are consist of lowercase letters a-z.

自行建立一個TrieNode的class，利用DFS來進行backtracking操作，途中已經掃描過的grid先利用#字元來暫時標記。helper function只要遇到prefix不符合的就停止DFS下去。

class TrieNode {
public:
    vector<TrieNode*> next;
    string word;
    TrieNode() : word(""), next(vector<TrieNode*>(26, nullptr)){}
    ~TrieNode() {
        for (TrieNode*& child : next) {
            if (child) delete child;
        }
    }
};
TrieNode* buildTrie(vector<string>& words) {
    TrieNode* root = new TrieNode();
    for (string& w : words) {
        TrieNode* cur = root;
        for (char c : w) {
            int idx = c - 'a';
            if (!cur->next[idx]) cur->next[idx] = new TrieNode();
            cur = cur->next[idx];
        }
        cur->word = w;
    }
    return root;
}
void helper(vector<vector<char> >& board, int i, int j, TrieNode* cur, vector<string>& res) {
    if (i < 0 || i >= board.size() || j < 0 || j >= board[0].size() || board[i][j] == '#' || !cur->next[board[i][j] - 'a']) return;
    char c = board[i][j];
    cur = cur->next[c - 'a'];
    if (!cur->word.empty()) {
        res.push_back(cur->word);
        cur->word.clear();
    }
    board[i][j] = '#';
    helper(board, i - 1, j, cur, res);
    helper(board, i + 1, j, cur, res);
    helper(board, i, j - 1, cur, res);
    helper(board, i, j + 1, cur, res);
    board[i][j] = c;
}
vector<string> findWords(vector<vector<char>>& board, vector<string>& words) { // time: (m * n * 4^(average word length)); space: O(4^(average word length))
    vector<string> res;
    if (board.empty() || board[0].empty() || words.empty()) return res;
    TrieNode* root = buildTrie(words);
    for (int i = 0; i < board.size(); ++i) {
        for (int j = 0; j < board[0].size(); ++j) {
            helper(board, i, j, root, res);
        }
    }
    return res;
}