在python中反转字符串中的子字符串
我正在编写一个程序来反转python中括在括号中的子字符串。结果字符串不应包含任何括号。我打印b1、b2和ch用于测试目的。似乎在while循环内for循环的第二次迭代中,b1变量没有使用正确的索引进行更新。 我试着写一个如下的解决方案:在python中反转字符串中的子字符串,python,python-3.x,substring,Python,Python 3.x,Substring,我正在编写一个程序来反转python中括在括号中的子字符串。结果字符串不应包含任何括号。我打印b1、b2和ch用于测试目的。似乎在while循环内for循环的第二次迭代中,b1变量没有使用正确的索引进行更新。 我试着写一个如下的解决方案: def reverseParentheses(s): r = s sstring = '' astring = '' b1 = b2 = 0 count = 0 for ch in s: if c
def reverseParentheses(s):
r = s
sstring = ''
astring = ''
b1 = b2 = 0
count = 0
for ch in s:
if ch == '(':
count+=1
elif ch ==')':
count+=1
else:
pass
while True:
b1 = b2 = 0
for ch in r:
if ch == '(':
b1 = r.index(ch)
print("b1= ",b1, ch)
if ch == ')':
b2 = r.index(ch)
print("b2= ",b2, ch)
sstring = r[b2-1:b1:-1]
print(r)
print(sstring)
astring = r[0:b1]+sstring+r[b2+1:]
print(astring)
r = astring
break
if len(astring)+count == len(s):
break
return r
s = "a(bcdefghijkl(mno)p)q"
print(reverseParentheses(s))
这是我得到的输出:
aonmpbcdefghijklq
这是我期望的输出:
apmnolkjihgfedcbq处理嵌套分隔符的一个好方法是使用堆栈。遇到开头分隔符时,将新集合推送到堆栈中<代码>弹出()。这将保持嵌套顺序正确 这里有一种方法可以做到这一点(它不检查平衡括号,但不难添加):
一种方法,通过找到括号的位置并从内到外反转(因此包含在偶数个括号之间的括号保持不变),最后去掉括号:
s = "a(bcdefghijkl(mno)p)q"
leftp = reversed([pos for pos, char in enumerate(s) if char == "("])
rightp = [pos for pos, char in enumerate(s) if char == ")"]
for i in zip(leftp,rightp):
subs = s[i[0]+1:i[1]][::-1]
s = s[:i[0]+1]+subs+s[i[1]:]
for c in ["(", ")"]:
s = s.replace(c, "")
print(s) # Outputs "apmnolkjihgfedcbq"
编辑
对于非嵌套的括号,正如所指出的那样。@Mark Meyer,您可以按描述找到它们,并且适用相同的规则
def find_parens(s):
toret = {}
pstack = []
for i, c in enumerate(s):
if c == '(':
pstack.append(i)
elif c == ')':
if len(pstack) == 0:
raise IndexError("No matching closing parens at: " + str(i))
toret[pstack.pop()] = i
if len(pstack) > 0:
raise IndexError("No matching opening parens at: " + str(pstack.pop()))
return toret
s = "a(bcd)efghijkl(mno)pq"
parens = find_parens(s)
for leftp, rightp in parens.items():
subs = s[leftp+1:rightp][::-1]
s = s[:leftp+1]+subs+s[rightp:]
for c in ["(", ")"]:
s = s.replace(c, "")
print(s) # Outputs "adcbefghijklonmpq"
你期望的输出是什么?它输出这个:b1=1(b1=1(b2=17)a(bcdefghijkl(mno)p)q onm(lkjihgfedcb aonm(lkjihgfedcbp)q b1=4(b2=17)aonm(lkjihgfedcbp)q pbcdefghijkl aonmpbcdefghijklq aonmpbcdefghijklq。最后一行是程序的最终结果。在最深嵌套的paren级别上查找parens索引,在中间反转字符,重复,直到没有parens剩余。不要在注释中回答,您的问题包括预期输出和实际输出是的,这是对问题的正确评估。似乎字符串有问题,例如:
s=“(ab)cdefghijklmno(pq)”
-->“conmlkjihgfedcbadefghijklmnopq”比您现在的更健壮hehe:PSeems现在非常防弹。
def find_parens(s):
toret = {}
pstack = []
for i, c in enumerate(s):
if c == '(':
pstack.append(i)
elif c == ')':
if len(pstack) == 0:
raise IndexError("No matching closing parens at: " + str(i))
toret[pstack.pop()] = i
if len(pstack) > 0:
raise IndexError("No matching opening parens at: " + str(pstack.pop()))
return toret
s = "a(bcd)efghijkl(mno)pq"
parens = find_parens(s)
for leftp, rightp in parens.items():
subs = s[leftp+1:rightp][::-1]
s = s[:leftp+1]+subs+s[rightp:]
for c in ["(", ")"]:
s = s.replace(c, "")
print(s) # Outputs "adcbefghijklonmpq"