【leetcode】999. 可以被一步捕获的棋子数

int numRookCaptures(char** board, int boardSize, int* boardColSize){
    int cnt = 0, st = 0, ed = 0;
        int dx[4] = {0, 1, 0, -1};
        int dy[4] = {1, 0, -1, 0};
        for (int i = 0; i < 8; ++i) {
            for (int j = 0; j < 8; ++j) {
                if (board[i][j] == ‘R‘) {
                    st = i;
                    ed = j;
                    break;
                }
            }
        }
        for (int i = 0; i < 4; ++i) {
            for (int step = 0;; ++step) {
                int tx = st + step * dx[i];
                int ty = ed + step * dy[i];
                if (tx < 0 || tx >= 8 || ty < 0 || ty >= 8 || board[tx][ty] == ‘B‘) {
                    break;
                }
                if (board[tx][ty] == ‘p‘) {
                    cnt++;
                    break;
                }
            }
        }
        return cnt;

}