第四题字符串代码: 通过80%超时,O(n)KMP做法,是python太慢了吗? n = 9 mo = "abaabaabaab" m = 3 m_str = ["aba","ab","abaaba"] n = len(mo) next = [0 for i in range(n+1)] def getnext(n2): i = 0 j = -1 while i<n2: if j==-1 or mo[i] == mo[j]: i+=...