你的示例和斐波那契函数都可以在不使用堆栈的情况下进行迭代重写。
这里有一个需要使用堆栈的例子,阿克曼函数:
def ack(m, n):
assert m >= 0 and n >= 0
if m == 0: return n + 1
if n == 0: return ack(m - 1, 1)
return ack(m - 1, ack(m, n - 1))
消除递归:
def ack_iter(m, n):
stack = []
push = stack.append
pop = stack.pop
RETURN_VALUE, CALL_FUNCTION, NESTED = -1, -2, -3
push(m) # push function arguments
push(n)
push(CALL_FUNCTION) # push address
while stack: # not empty
address = pop()
if address is CALL_FUNCTION:
n = pop() # pop function arguments
m = pop()
if m == 0: # return n + 1
push(n+1) # push returned value
push(RETURN_VALUE)
elif n == 0: # return ack(m - 1, 1)
push(m-1)
push(1)
push(CALL_FUNCTION)
else: # begin: return ack(m - 1, ack(m, n - 1))
push(m-1) # save local value
push(NESTED) # save address to return
push(m)
push(n-1)
push(CALL_FUNCTION)
elif address is NESTED: # end: return ack(m - 1, ack(m, n - 1))
# old (m - 1) is already on the stack
push(value) # use returned value from the most recent call
push(CALL_FUNCTION)
elif address is RETURN_VALUE:
value = pop() # pop returned value
else:
assert 0, (address, stack)
return value
这里不需要在堆栈上放置CALL_FUNCTION
、RETURN_VALUE
标签和value
。
print(ack(2, 4)) # -> 11
print(ack_iter(2, 4))
assert all(ack(m, n) == ack_iter(m, n) for m in range(4) for n in range(6))
print(ack_iter(3, 4)) # -> 125