Question

这段代码应该带一个字符串并返回一个没有重复字符的字符串（不是＆＃34; aa＆＃34;，＆＃34; bb＆＃34;，＆＃34;＆＃34; 55＆＃34 ;等）。它通过递归工作：函数调用自身，细化字符串直到它没有重复，但然后它返回原始字符串。我无法理解如何修复它。代码如下，抱歉，如果这个问题太过琐碎，但对我来说不是。谢谢您的考虑。（并且抱歉＆＃34;黑客攻击＆＃34;标题格式化）

def removeRepetitions(s):
    result=""
    for i in range(len(s)-1):
        if len(s)!=1 and s[i]==s[i+1]:
            result=s[:i+1]+s[i+2:]
            removeRepetitions(result)
            return result

Answer 1

如果缩进正确，则返回值的缩进不正确，并且您没有分配递归函数的return。您还需要在开头指定result到s而不是空字符串。您的终端案例必须返回s中的单个字符：

def removeRepetitions(s):
    result = s
    for i in range(len(s)-1):
        if len(s)!=1 and s[i]==s[i+1]:
            result = removeRepetitions(s[:i+1]+s[i+2:])
    return result

>>> removeRepetitions('Mississippi')
'Misisipi'

但是你似乎在混合递归和迭代解决方案，更加递归的解决方案不需要for循环：

def removeRepetitions(s):
    if len(s) <= 1:
        return s
    c, s = s[0], removeRepetitions(s[1:])
    if c == s[0]:
        return s
    return c+s

>>> removeRepetitions('Mississippi')
'Misisipi'

Answer 2

你缺少的主要内容是返回陈述以涵盖所有可能性。您也不需要中间变量result。纠正缩进后：

def removeRepetitions(s):
    for i in range(len(s) - 1):
        if s[i] == s[i+1]:  # found a dup!
            # Assemble a return value by taking everything prior to the
            # duplicate (which is duplicate free) and concatenating the
            # result of removing duplicates from the remainder of the
            # input string, excluding the first of the duplicated chars
            return s[:i] + removeRepetitions(s[i+1:])
    return s  # return input string if we made it through w/o any duplicates

print(removeRepetitions("aaaabccdddeee"))    # => "abcde"

另请注意，对于最终的return语句，检查len(s)!=1是无关紧要的。

pProblem在函数中具有递归和范围（Python）

2 个答案: