我做了一个迷宫寻路递归功能作为家庭作业,但我遇到了一个问题。我知道我已经完成了所有事情(通过调试),路径被找到等等。但是当所有递归函数开始返回True时,原始/第一个不是!我无法找到使其返回true的方法(递归调用采用“if,true”格式)。
我希望你们能帮助我,对不好的英语感到抱歉,因为这不是我的母语。
*这里是代码:
def pathExists(labyrinth, currCoord, destination, previousRule, visited):
if currCoord == destination:
return True
if currCoord not in labyrinth:
print "False"
return False
rule = labyrinth[currCoord]
if rule == Any:
previousRule = rule
print currCoord
if (pathExists(labyrinth, (currCoord[0], currCoord[1] - 1), destination, rule, visited) or
pathExists(labyrinth, (currCoord[0] + 1, currCoord[1]), destination, rule, visited) or
pathExists(labyrinth, (currCoord[0] - 1, currCoord[1]), destination, rule, visited) or
pathExists(labyrinth, (currCoord[0], currCoord[1] + 1), destination, rule, visited)):
print "True"
return True
else:
print"outro"
elif rule == Bridge:
print currCoord
currCoord = nextCoord(currCoord, previousRule)
if pathExists(labyrinth, currCoord, destination, rule, visited):
print "True"
return True
else:
print currCoord
if currCoord in visited:
print "False"
return False
visited.append(currCoord)
previousRule = rule
currCoord = nextCoord(currCoord, rule)
if pathExists(labyrinth, currCoord, destination, rule, visited):
print "True"
return True
答案 0 :(得分:1)
没有所有代码我都错了,但我认为你需要一个global / class var来存储返回值。逻辑:第一次调用不会返回任何内容(或返回None),因为它正在调用其他函数
要明确:
a = [1,2,[4,6],3,4]
ret = None
def f(x):
global ret
if isinstance(x, list):
f(x[0])
else:
ret = x
return x
var = f(a)
print(var, ret)
如果你测试,你会看到var = None,但是ret是1(而不是全局我建议你在类中包含代码并使用像self.ret这样的东西)。
P.S。:正如Joel Cornett所说,使用英文/抽象var名称可以使所有代码更具可读性!