Given a 2D board and a word, find if the word exists in the grid. The word can 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. For example, Given board = [ ["ABCE"], ["SFCS"], ["ADEE"] ] word = "ABCCED", -> re…

leetcode_word_search.cpp

#define MAXN 1000

bool used[MAXN][MAXN];
int dir[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

bool valid(int i, int j, int n, int m) {
    return i >= 0 && i < n && j >= 0 && j < m;
}

bool find(vector<vector<char> >& board, bool used[MAXN][MAXN], int n, int m, 
        int i, int j, string& word, int kth) {
    if (kth == word.length())
        return true;
    bool ret = false;
    for (int k = 0; k < 4; ++k) {
        int next_i = i + dir[k][0];
        int next_j = j + dir[k][1];
        if (valid(next_i, next_j, n, m) 
                && !used[next_i][next_j] 
                && board[next_i][next_j] == word[kth]) {
            used[next_i][next_j] = true;
            ret = find(board, used, n, m, next_i, next_j, word, kth + 1);
            used[next_i][next_j] = false;
            if (ret)
                break;
        }
    }
    return ret;
}

class Solution {
public:
    bool exist(vector<vector<char> > &board, string word) {
        if (word.empty())
            return true;
        int n = board.size();
        int m = board[0].size();
        memset(used, false, sizeof(used));
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < m; ++j)
                if (board[i][j] == word[0]) {
                    used[i][j] = true;
                    if (find(board, used, n, m, i, j, word, 1))
                        return true;
                    used[i][j] = false;
                }
        return false;
    }
};