题目链接：79. 单词搜索 - 力扣（LeetCode）

要在一个二维数组里面找到一条单词路径，可以先遍历二维数组找到单词入口，然后往上下左右深度遍历，访问过的元素直接修改成字符串结束符，访问完改回去

class Solution {
public:
    string word;
    int row, column;
    vector<vector<char> > board;

    bool dfs(int i, int j, int count) {
        if (i >= row || j >= column || i < 0 || j < 0 || board[i][j] != word[count])
            return false;
        if (count == word.size() - 1)
            return true;
        board[i][j] = '\0';
        ++count;
        bool next = dfs(i - 1, j, count) || dfs(i, j - 1, count) || dfs(i + 1, j, count) || dfs(i, j + 1, count);
        board[i][j] = word[--count];
        return next;
    }

    bool exist(vector<vector<char> > &board, string word) {
        this->board = move(board);
        this->word = move(word);
        row = this->board.size();
        column = this->board[0].size();
        for (int i = 0; i < row; ++i)
            for (int j = 0; j < column; ++j) {
                if (dfs(i, j, 0))
                    return true;
            }
        return false;
    }
};