python - 递归删除偶数相邻重复字母的代码

Question

最近我被要求编写代码以在实时编码测试中递归删除偶数的相邻字母。当时我无法编写代码。之后我尽力这样做。我的问题是什么逻辑和代码？

例如 cbbbaaaabbbccc => cbbbbbbccc => cccc => 空字符串

例如 aabbc => bbc => c

例如 abbbccc => abbbccc - 因为偶数中没有字母重复

编辑 - 我根据 Rory 的建议编辑了代码，但仍然不明白为什么在字符串变空后它会进入递归调用而不是跳出循环

str = "cbbbaaaabbbccc"   


def remUtil(str):
i = 0
ind = 0
while i < (len(str)-1):
    j = 0
    if str[i] == str[i + 1]:
        j = i
        ind = i
        count = 0
        while j < len(str)-1 and str[j] == str[j + 1]:
            count = count + 1
            j = j + 1
            i = i + 1
        # if the no. of comparisons are odd then even letters compared
        if count % 2 != 0:
            str = str[:(ind)] + str[(ind + count) + 1:]
            #print(str)
            remUtil(str)


    else:
        i = i + 1

Answer 1

一个较短的版本，避免了对索引的操作:

def rm_even_duplicates(s):
    ls = list(s) + [None]
    tmp = [ls[0]]
    out = []
    for c in ls[1:]:
        if c != tmp[0]:
            if len(tmp) % 2 == 1:
                out.extend(tmp)
            tmp = []
        tmp.append(c)
    # The recursive part, if you want to do it that way;
    # that could as well have been a while loop
    if len(out) == len(s):
        return ''.join(out)
    else:
        return rm_even_duplicates(out)

一些示例和您的测试用例:

print(rm_even_duplicates('aaabbcdddd'))
# aaac
print(rm_even_duplicates('aaabbccaaadda'))
# aaaaaaa

assert rm_even_duplicates('cbbbaaaabbbccc') == ''
assert rm_even_duplicates('aabbc') == 'c'
assert rm_even_duplicates('abbbccc') == 'abbbccc'