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.

For example,
Given words = [“oath”,”pea”,”eat”,”rain”] and board =

[
  ['o','a','a','n'],
  ['e','t','a','e'],
  ['i','h','k','r'],
  ['i','f','l','v']
]

Return [“eat”,”oath”].

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

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引用Trie类实现

//word search II
class Solution {
public:
    void bt_search(vector<vector<char>> &board, vector<vector<bool>> &used, 
        int x, int y, string word){
        if(x<0 || y<0 || x==board.size() || y==board[0].size() || used[x][y]){
            return;
        }
        word += board[x][y];
        if(!trie.startsWith(word)){
            return;
        }
        if(trie.search(word)){
            result.push_back(word);
        }
        used[x][y] = true;
        bt_search(board, used, x-1, y, word);
        bt_search(board, used, x, y-1, word);
        bt_search(board, used, x+1, y, word);
        bt_search(board, used, x, y+1, word);
        used[x][y] = false;
    }

    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        for(auto w=words.begin(); w!=words.end(); ++w){
            trie.insert(*w);
        }
        vector<vector<bool>> used(board.size(), vector<bool>(board[0].size(), false));
        for(int i=0; i<board.size(); i++){
            for(int j=0; j<board[i].size(); j++){
                bt_search(board, used, i, j, "");
            }
        }
        sort(result.begin(), result.end());
        auto it = unique(result.begin(), result.end());
        result.resize(distance(result.begin(), it));
        return result;
    }
private:
    Trie trie;
    vector<string > result;
};