9-16每日一题：单词搜索二

Leetcode212. 单词搜索 II

这道题需要用的前缀树，所以先做一道预备题：208. 实现 Trie (前缀树)

class Trie {
public:
    struct Node {
        bool is_end;
        Node *son[26];
        Node () {
            is_end = false;
            for (int i = 0; i < 26; i ++ )
                son[i] = NULL;
        }
    }*root;
    /** Initialize your data structure here. */
    Trie() {
        root = new Node();
    }
    
    /** Inserts a word into the trie. */
    void insert(string word) {
        auto p = root;
        for (auto c : word) {
            int u = c - 'a';
            if (!p->son[u]) p->son[u] = new Node();
            p = p->son[u];
        }
        p->is_end = true;
    }
    
    /** Returns if the word is in the trie. */
    bool search(string word) {
        auto p = root;
        for (auto& c : word) {
            int u = c - 'a';
            if (!p->son[u]) return false;
            p = p->son[u];
        }
        return p->is_end;
    }
    
    /** Returns if there is any word in the trie that starts with the given prefix. */
    bool startsWith(string word) {
        auto p = root;
        for (auto& c : word) {
            int u = c - 'a';
            if (!p->son[u]) return false;
            p = p->son[u];
        }
        return true;
    }
};

/**
 * Your Trie object will be instantiated and called as such:
 * Trie* obj = new Trie();
 * obj->insert(word);
 * bool param_2 = obj->search(word);
 * bool param_3 = obj->startsWith(prefix);
 */

思路：哈希+前缀树+回溯

class Solution {
public:
    struct Node {
        int id;
        Node *son[26];

        Node () {
            id = -1;
            for (int i = 0; i < 26; i ++ )
                son[i] = NULL;
        }
    }*root;

    vector<vector<char>> g;
    unordered_set<int> ids;
    int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};

    void insert(string& word, int id) {
        auto p = root;
        for (auto& c : word) {
            int u = c - 'a';
            if (!p->son[u]) p->son[u] = new Node();
            p = p->son[u];
        }
        p->id = id;
    }

    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        g = board;
        root = new Node();
        for (int i = 0; i < words.size(); i ++ ) insert(words[i], i);

        for (int i = 0; i < g.size(); i ++ )
            for (int j = 0; j < g[0].size(); j ++ ) {
                int u = g[i][j] - 'a';
                if (root->son[u]) 
                    dfs(i, j, root->son[u]);
            }
        vector<string> res;
        for (auto id : ids) res.push_back(words[id]);
        return res;
    }

    void dfs(int x, int y, Node* p) {
        if (p->id != -1) {
            ids.insert(p->id);
        }
        char t = g[x][y];
        g[x][y] = '.';
        for (int i = 0; i < 4; i ++ ) {
            int a = x + dx[i], b = y + dy[i];
            if (a >= 0 && a < g.size() && b >= 0 && b < g[0].size() && g[a][b] != '.') {
                int u = g[a][b] - 'a';
                if (p->son[u]) dfs(a, b, p->son[u]);
            }
        }
        g[x][y] = t;
    }
};