[LeetCode] Word Search II 词语搜索之二

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:

board = [

  ['o','a','a','n'],

  ['e','t','a','e'],

  ['i','h','k','r'],

  ['i','f','l','v']

]

words = ["oath","pea","eat","rain"]

Output: ["eat","oath"]

Note:

All inputs are consist of lowercase letters a-z.
The values of words are distinct.

You would need to optimize your backtracking to pass the larger test. Could you stop backtracking earlier?

If the current candidate does not exist in all words' prefix, you could stop backtracking immediately. What kind of data structure could answer such query efficiently? Does a hash table work? Why or why not? How about a Trie? If you would like to learn how to implement a basic trie, please work on this problem: Implement Trie (Prefix Tree) first.

这道题是在之前那道 Word Search 的基础上做了些拓展，之前是给一个单词让判断是否存在，现在是给了一堆单词，让返回所有存在的单词，在这道题最开始更新的几个小时内，用 brute force 是可以通过 OJ 的，就是在之前那题的基础上多加一个 for 循环而已，但是后来出题者其实是想考察字典树的应用，所以加了一个超大的 test case，以至于 brute force 无法通过，强制我们必须要用字典树来求解。LeetCode 中有关字典树的题还有 Implement Trie (Prefix Tree) 和 Add and Search Word - Data structure design，那么我们在这题中只要实现字典树中的 insert 功能就行了，查找单词和前缀就没有必要了，然后 DFS 的思路跟之前那道 Word Search 基本相同，请参见代码如下：

class Solution {

public:

    struct TrieNode {

        TrieNode *child[];

        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<string> findWords(vector<vector<char>>& board, vector<string>& words) {

        vector<string> res;

        if (words.empty() || board.empty() || board[].empty()) return res;

        vector<vector<bool>> visit(board.size(), vector<bool>(board[].size(), false));

        Trie T;

        for (auto &a : words) T.insert(a);

        for (int i = ; i < board.size(); ++i) {

            for (int j = ; 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<vector<char>>& board, TrieNode* p, int i, int j, vector<vector<bool>>& visit, vector<string>& res) {

        if (!p->str.empty()) {

            res.push_back(p->str);

            p->str.clear();

        }

        int d[][] = {{-, }, {, }, {, -}, {, }};

        visit[i][j] = true;

        for (auto &a : d) {

            int nx = a[] + i, ny = a[] + j;

            if (nx >=  && nx < board.size() && ny >=  && ny < board[].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;

    }

};