I want to replace my function foo with foo2(non-resursion), but foo2 works incorrect.
What's wrong with foo2?
def foo(n, k=0,s=0):
if k < n:
for i in xrange(k==0,10):
foo(n, k+1, 10*s + i)
else:
print s,
def foo2(n):
s=0
for k in xrange(n):
st = s
for i in xrange(k==0, 10):
st = 10* st + i
print st
foo(3)
foo2(3)
Updated
If I replace 10*s + i with s + i**3, How can I rewrite it?
foo prints 10n-1 ~ 10n-1; Iterate xrange(10**(n-1), 10**n).
def foo2(n):
for s in xrange(10**(n-1), 10**n):
print s,
Following is a translation of the recursive funciton using stack:
def foo2(n):
stack = [(0, 0)] # corresponding to (..., k=0, s=0)
while stack:
k, s = stack.pop(0)
if k < n:
for i in xrange(k==0, 10):
stack.append((k+1, 10*s + i))
else:
print s,
NOTE To implement strictly equivalent iterative version, you should also push iterator (xrange...); consume only one item at a time in a loop.
