回溯法。

class Solution {
public:
    int n, m;
    int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
    
    bool check(int x, int y) {
        return x >= 0 && x < n && y >= 0 && y < m;
    }
    
    bool dfs(int x, int y, int idx, string &word, vector<vector<char>> &board) {
        if (word[idx] != board[x][y]) return false;
        if (idx == word.size() - 1) return true;
        char t = board[x][y];
        board[x][y] = ' ';
        for (int i = 0; i < 4; i++) {
            int a = x + dx[i], b = y + dy[i];
            if (check(a, b))
                if (dfs(a, b, idx + 1, word, board))
                    return true;
        }
        board[x][y] = t;  // 回溯
        return false;
    }
    
    bool exist(vector<vector<char>>& board, string &word){
        n = board.size(), m = board[0].size();
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                if (dfs(i, j, 0, word, board))
                    return true;
        return false;
    }
};