Question

我正在编写一个程序来反转python括号中的子字符串。结果字符串不应包含任何括号。我正在打印b1，b2和ch以进行测试。似乎在while循环内for循环的第二次迭代中，未使用正确的索引更新b1变量。我试图编写如下解决方案：

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： on 这是我期望的输出： apmnolkjihgfedcbq

Answer 1

处理嵌套定界符的一种好方法是使用堆栈。当遇到分隔符时，将新集合推入堆栈。 pop()，当您找到结案时。这样可以保持正确的嵌套顺序。

这是执行此操作的一种方法（它不检查括号是否平衡，但是添加起来并不难）：

s = "a(bcdefghijkl(mno)p)q"
stack = [[]] # accumulate letters in stack[0]
for l in s:
    if l == '(':
        stack.append([])        # start a new level
    elif l == ')':
        sub = stack.pop()[::-1] # pop the last level and reverse
        stack[-1].extend(sub)   # add to current 
    else:
        stack[-1].append(l)     # add to current

''.join(stack[0]) #'apmnolkjihgfedcbq'

Answer 2

一种找到括号位置并从里到外反转的方法（这样，包含在偶数个括号之间的那些保持不变），最后摆脱了括号：

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。@ Mark Meyer指出的未嵌套的括号，您可以按照here的说明找到它们，并应用相同的规则

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"

在Python中反转字符串中的子字符串

2 个答案: