def have_five(self, current_i, current_j): #四个方向计数 横 竖 左斜 右斜 hcount = 1 vcount = 1 lbhcount = 1 rbhcount = 1 temp = ChessboardState.EMPTY j in range(current_j - 1, -1, -1): #横向往左 from (current_j - 1) to 0 temp = self.__chessMap[current_i][j] if temp == ChessboardState.EMPTY or temp != self.__currentState: break hcount = hcount + 1 j in range(current_j + 1, N): #横向往右 from (current_j + 1) to N temp = self.__chessMap[current_i][j] if temp == ChessboardState.EMPTY or temp != self.__currentState: break hcount = hcount + 1 hcount >= 5: return True
i in range(current_i - 1, -1, -1): # from (current_i - 1) to 0 temp = self.__chessMap[i][current_j] if temp == ChessboardState.EMPTY or temp != self.__currentState: break vcount = vcount + 1 i in range(current_i + 1, N): # from (current_i + 1) to N temp = self.__chessMap[i][current_j] if temp == ChessboardState.EMPTY or temp != self.__currentState: break vcount = vcount + 1 vcount >= 5: return True
i, j in zip(range(current_i - 1, -1, -1), range(current_j - 1, -1, -1)): temp = self.__chessMap[i][j] if temp == ChessboardState.EMPTY or temp != self.__currentState: break lbhcount = lbhcount + 1 i, j in zip(range(current_i + 1, N), range(current_j + 1, N)): temp = self.__chessMap[i][j] if temp == ChessboardState.EMPTY or temp != self.__currentState: break lbhcount = lbhcount + 1 lbhcount >= 5: return True
i, j in zip(range(current_i - 1, -1, -1), range(current_j + 1, N)): temp = self.__chessMap[i][j] if temp == ChessboardState.EMPTY or temp != self.__currentState: break rbhcount = rbhcount + 1 i, j in zip(range(current_i + 1, N), range(current_j - 1, -1, -1)): temp = self.__chessMap[i][j] if temp == ChessboardState.EMPTY or temp != self.__currentState: break rbhcount = rbhcount + 1 rbhcount >= 5: return True

这样是不是就写完了，五子棋的逻辑全部实现~ 

NO，别高兴得太早，我想说，我好恶心，上面那个代码，简直丑爆了，再看一眼，重复的写了这么多for，这么多if，这么多重复的代码块，让我先去吐会儿……

好了，想想办法怎么改，至少分了4根轴，是重复的对不对，然后每根轴分别从正负两个方向去统计，最后加起来，两个方向，也是重复的对不对。

于是我们能不能只写一个方向的代码，分别调2次，然后4根轴，分别再调4次，2*4=8，一共8行代码搞定试试。

因为有45°和135°这两根斜轴的存在，所以方向上应该分别从x和y两个轴来控制正负，于是可以这样，先写一个函数，按照方向来统计：

xdirection=0,ydirection=1       表示从y轴正向数；

xdirection=0,ydirection=-1     表示从y轴负向数；

xdirection=1,ydirection=1       表示从45°斜轴正向数；

……

不一一列举了，再加上边界条件的判断，于是有了以下函数：

def count_on_direction(self, i, j, xdirection, ydirection, color): count = 0 xdirection != 0 and (j + xdirection * step < 0 or j + xdirection * step >= N): break if ydirection != 0 and (i + ydirection * step < 0 or i + ydirection * step >= N): break if self.__chessMap[i + ydirection * step][j + xdirection * step] == color: count += 1 else: break return count

于是乎，前面的have_five稍微长的好看了一点，可以变成这样：

def have_five(self, i, j, color): #四个方向计数 横 竖 左斜 右斜 hcount = 1 vcount = 1 lbhcount = 1 rbhcount = 1 hcount += self.count_on_direction(i, j, -1, 0, color) hcount += self.count_on_direction(i, j, 1, 0, color) if hcount >= 5: return True vcount += self.count_on_direction(i, j, 0, -1, color) vcount += self.count_on_direction(i, j, 0, 1, color) if vcount >= 5: return True lbhcount += self.count_on_direction(i, j, -1, 1, color) lbhcount += self.count_on_direction(i, j, 1, -1, color) if lbhcount >= 5: return True rbhcount += self.count_on_direction(i, j, -1, -1, color) rbhcount += self.count_on_direction(i, j, 1, 1, color) if rbhcount >= 5: return True