Question

我必须编写扫雷。我已经找到了一种方法来检查相邻的指数来计算当前指数周围有多少炸弹而不超出界限。但它很长，很丑，很可能效率低下。它们只是每个可能触发范围的唯一单元格的一系列条件语句。我的问题是，有更短的方法吗？并且实现一维扭曲阵列会使它变得更容易/更难吗？

int count = 0;
for (int i = 0; i < WIDTH; i++){
    for (int j = 0; j < HEIGHT; j++){
        **// index 0,0**
        if (i == 0 && j == 0 && field[i][j] != 10){
            if (field[i][j + 1] == 10)
                count++;
            if (field[i + 1][j] == 10)
                count++;
            if (field[i + 1][j + 1] == 10)
                count++;
        }
        **// index 0,9**
        else if (i == 0 && j == HEIGHT - 1 && field[i][j] != 10){
            if (field[i][j - 1] == 10)
                count++;
            if (field[i + 1][j - 1] == 10)
                count++;
            if (field[i + 1][j] == 10)
                count++;
        }

        **// index 9,0**
        else if (i == WIDTH - 1 && j == 0 && field[i][j] != 10){
            if (field[i - 1][j] == 10)
                count++;
            if (field[i][j + 1] == 10)
                count++;
            if (field[i - 1][j + 1] == 10)
                count++;
        }

        **// index 9,9**
        else if (i == WIDTH - 1 && j == HEIGHT - 1 && field[i][j] != 10){
            if (field[i - 1][j] == 10)
                count++;
            if (field[i][j - 1] == 10)
                count++;
            if (field[i - 1][j - 1] == 10)
                count++;
        }

        **// if first row**
        else if (i == 0 && (j != 0 && j != HEIGHT - 1) && field[i][j] != 10){
            if (field[i][j - 1] == 10)
                count++;
            if (field[i + 1][j - 1] == 10)
                count++;
            if (field[i + 1][j] == 10)
                count++;
            if (field[i + 1][j + 1] == 10)
                count++;
            if (field[i][j + 1] == 10)
                count++;
        }

        **// if last row**
        else if (i == WIDTH - 1 && (j != 0 && j != HEIGHT - 1) && field[i][j] != 10){
            if (field[i][j - 1] == 10)
                count++;
            if (field[i - 1][j - 1] == 10)
                count++;
            if (field[i - 1][j] == 10)
                count++;
            if (field[i - 1][j + 1] == 10)
                count++;
            if (field[i][j + 1] == 10)
                count++;
        }

        **// if first col**
        else if (j == 0 && (i != 0 && i != HEIGHT - 1) && field[i][j] != 10){
            if (field[i - 1][j] == 10)
                count++;
            if (field[i - 1][j + 1] == 10)
                count++;
            if (field[i][j + 1] == 10)
                count++;
            if (field[i + 1][j] == 10)
                count++;
            if (field[i + 1][j + 1] == 10)
                count++;
        }

        **// if last col**
        else if (j == HEIGHT - 1 && (i != 0 && i != WIDTH - 1) && field[i][j] != 10){
            if (field[i - 1][j - 1] == 10)
                count++;
            if (field[i - 1][j] == 10)
                count++;
            if (field[i][j - 1] == 10)
                count++;
            if (field[i + 1][j - 1] == 10)
                count++;
            if (field[i + 1][j] == 10)
                count++;
        }

        **// Cells that are fully surrounded**
        else if (field[i][j] != 10){
            if (field[i - 1][j - 1] == 10)
                count++;
            if (field[i - 1][j] == 10)
                count++;
            if (field[i - 1][j + 1] == 10)
                count++;
            if (field[i][j - 1] == 10)
                count++;
            if (field[i][j + 1] == 10)
                count++;
            if (field[i + 1][j - 1] == 10)
                count++;
            if (field[i + 1][j] == 10)
                count++;
            if (field[i + 1][j + 1] == 10)
                count++;
        }


        if (field[i][j] != 10)
            field[i][j] = count;
        count = 0;
    }
}

Answer 1

有一种标准方法可以处理这种情况 - 使用有效的移动定义一个数组并迭代它。以下是如何迭代8个相邻单元格的示例：

static int moves[8][2] = {{-1,0}, {-1, -1}, {0, -1}, {1, -1}, {1, 0}, {1, 1}, {0, 1}, {-1, 1}};
for (int i = 0; i < WIDTH; i++) {
    for (int j = 0; j < HEIGHT; j++) {
      for (int l = 0; l < 8 /*sizeof(move) / sizeof(move[0]) */; ++l) {
        int ti = i + move[l][0];
        int tj = j + move[l][1];
        if (ti < 0 || ti >= WIDTH || tj < 0 || tj >= HEIGHT) {
          continue;
        }
        // cell (ti, tj) is valid neighbor do something with it.
      }
    }
}

检查二维数组中的相邻索引

1 个答案: