[英]Recursively checking for balanced string in Python
I've been stuck on this for quite a while, I can't come up with recursive cases, in particular I don't understand how to split a list to check if its balanced.我已经坚持了很长一段时间,我无法提出递归案例,特别是我不明白如何拆分列表以检查其是否平衡。 If someone could help me I would really appreciate it.
如果有人可以帮助我,我将不胜感激。
def balanced_str(s):
"""
Return whether string s is balanced or not. A balanced string is one where
the string contains no parentheses
>>> balanced_str('a')
True
>>> balanced_str('abbcsi')
True
>>> balanced_str('ak)')
False
>>> balanced_str('hah(dh')
False
>>> balanced_str('()')
True
>>> balanced_str('(hghghgh)')
True
>>> balanced_str('((a))')
True
>>> balanced_str('((hahsh))')
True
>>> balanced_str('(gfjf)h)')
False
>>> balanced_str('(hhg)(hfhg)')
True
"""
if '(' not in s and ')' not in s:
return True
elif '(' in s and ')' not in s or ')' in s and '(' not in s:
return False
else:
if s[0] == '(' and s[len(s) - 1] == ')':
return balanced_str(s[1:len(s) - 2])
A simple, iterative approach could be to create a tiny lexer.一种简单的迭代方法可能是创建一个小型词法分析器。 It will increase a counter
o
when an opening parentheses (
appears, and decreases the counter if a closing parentheses )
appears.当出现左括号
(
出现,如果出现右括号,则减少计数器)
时,它将增加计数器o
。 If meanwhile searching the string the counter o
gets negative, or by the end of the loop the counter o
is not zero, the test fails:如果同时搜索字符串计数器
o
负,或者在循环结束时计数器o
不为零,则测试失败:
def balanced_str(s):
o = 0
for c in s:
if c == ')':
if o <= 0:
# this only happens if there are more closing
# parentheses then opening parentheses.
return False
o -= 1
elif c == '(':
o += 1
# all parentheses should be closed
return o == 0
For your test cases this approach works:对于您的测试用例,这种方法有效:
>>> balanced_str('a')
True
>>> balanced_str('abbcsi')
True
>>> balanced_str('ak)')
False
>>> balanced_str('hah(dh')
False
>>> balanced_str('()')
True
>>> balanced_str('(hghghgh)')
True
>>> balanced_str('((a))')
True
>>> balanced_str('((hahsh))')
True
>>> balanced_str('(gfjf)h)')
False
>>> balanced_str('(hug)(hfhg)')
True
For a recursive approach, you can create a small helper function that takes more parameters (ie. the number of parens we've seen so far).对于递归方法,您可以创建一个带有更多参数的小型辅助函数(即我们目前看到的括号数量)。 Below this approach, you can see how you can do it without a helper function through the use of
global
在这种方法下面,您可以看到如何在没有辅助函数的情况下通过使用
global
def balanced_str(s):
"""
Return whether string s is balanced or not. A balanced string is one where
the string contains no parentheses
>>> balanced_str('a')
True
>>> balanced_str('abbcsi')
True
>>> balanced_str('ak)')
False
>>> balanced_str('hah(dh')
False
>>> balanced_str('()')
True
>>> balanced_str('(hghghgh)')
True
>>> balanced_str('((a))')
True
>>> balanced_str('((hahsh))')
True
>>> balanced_str('(gfjf)h)')
False
>>> balanced_str('(hhg)(hfhg)')
True
"""
return helper(s,0)
def helper(s, numP):
if len(s)==0: return numP==0
if numP < 0: return False
if s[0] == "(": return helper(s[1:], numP+1)
elif s[0] == ")": return helper(s[1:], numP-1)
return helper(s[1:], numP)
Without helper:没有帮手:
def balanced_str(s):
"""
Return whether string s is balanced or not. A balanced string is one where
the string contains no parentheses
>>> balanced_str('a')
True
>>> balanced_str('abbcsi')
True
>>> balanced_str('ak)')
False
>>> balanced_str('hah(dh')
False
>>> balanced_str('()')
True
>>> balanced_str('(hghghgh)')
True
>>> balanced_str('((a))')
True
>>> balanced_str('((hahsh))')
True
>>> balanced_str('(gfjf)h)')
False
>>> balanced_str('(hhg)(hfhg)')
True
"""
try:
numP
except NameError:
numP = 0
global numP
if len(s)==0: return numP==0
if numP < 0: return False
if s[0] == "(":
numP += 1
return balanced_str(s[1:])
elif s[0] == ")":
numP -= 1
return balanced_str(s[1:])
return balanced_str(s[1:])
Here's my candidate solution:这是我的候选解决方案:
def balanced(iterable, semaphore=0):
if semaphore < 0 or len(iterable) == 0:
return semaphore == 0
first, *rest = iterable
return balanced(rest, semaphore + { "(": 1, ")": -1 }.get(first, 0))
I've renamed balanced_str()
to balanced()
since if it's written properly, it should handle strings or lists of characters (ie iterables ):我已将
balanced_str()
重命名为balanced()
因为如果它写得正确,它应该处理字符串或字符列表(即iterables ):
>>> balanced('a')
True
>>> balanced(['a', 'b', 'b', 'c', 's', 'i'])
True
>>> balanced('ak)')
False
>>> balanced(['h', 'a', 'h', '(', 'd', 'h'])
False
>>> balanced('()')
True
>>> balanced(['(', 'h', 'g', 'h', 'g', 'h', 'g', 'h', ')'])
True
>>> balanced('((a))')
True
>>> balanced(['(', '(', 'h', 'a', 'h', 's', 'h', ')', ')'])
True
>>> balanced('(gfjf)h)')
False
>>> balanced(['(', 'h', 'h', 'g', ')', '(', 'h', 'f', 'h', 'g', ')'])
True
I believe this is true of other proposed solutions as well, not just mine.我相信其他提议的解决方案也是如此,而不仅仅是我的。
Here's a recursive approach.这是一种递归方法。 I think this is pretty concise and intuitive
我认为这非常简洁直观
def is_balanced(s, c=0):
if len(s) == 0:
if c == 0:
return True
else:
return False
elif s[0] == "(":
c = c + 1
return is_balanced(s[1 : len(s)], c)
elif s[0] == ")":
c = c - 1
return is_balanced(s[1 : len(s)], c)
else:
return is_balanced(s[1 : len(s)], c)
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