如何仅计算给定字符串中的连续重复子字符串

Question

def count_overlapping(sequence, sub):
    counts = 0
    n = len(sub)
    while sub in sequence:
        counts += 1
        sequence = sequence[(sequence.find(sub) + n-1):]
    return counts

输入：sequence=agatabatagataagataagata

输入：sub=agata

产出：3

这是要求的输出，但我的程序给出了4。我如何忽略非重复的输出

有人请带我到这里来

Answer 1

下面是一种基于正则表达式的方法，它使用re.sub删除子字符串的所有重复组。然后，为了找到存在的子串的数量，我们只需将长度差除以子串的长度

sequence = "agatabttagataagataagatagatabagata"
out = re.sub(r'(?:agata){2,}', '', sequence)
num = (len(sequence) - len(out)) / len('agata')
print(num)

这将打印：3

Answer 2

最简单但效率不高的解决方案是每次将子字符串相乘，直到在字符串中再也找不到它，然后找到最大重复次数：

s = "agatabttagataagataagatagatabagata"
sub = "agata"

counts = 0
while sub * (counts+1) in s:
    counts += 1

print(counts)

这就产生了3