|
currentState:
break
hcount = hcount + 1
#H-结果
if hcount >= 5:
return True
#V-上
for 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
#V-下
for 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
#V-结果
if vcount >= 5:
return True
#LB-上
for 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
#LB-下
for 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
#LB-结果
if lbhcount >= 5:
return True
#RB-上
for 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
#RB-下
for 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
#LB-结果
if 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
for step in range(1, 5): #除当前位置外,朝对应方向再看4步
if 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 竖 横 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
还是一大排重复的代码呀，我还是觉得它丑啊，我真的不是处女座，但是这个函数是真丑啊，能不能让它再帅一点，当然可以，4个重复块再收成一个函数，循环调4次，是不是可以，好，就这么干，于是have_five就又漂亮了一点点：
def have_five(self, i, j, color):
#四个方向计数 竖 横 左斜 右斜
directions = [[(- |