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字符串模板:字符串匹配

1 kmp 1 2 3 4 5 6 7 8 def kmp(s): nxt, i = [-1], -1 for c in s: while i>=0 and s[i]!=c: i = nxt[i] i += 1 nxt.append(i) return nxt # nxt[i]:i-1结尾的最大真前缀长度 2 manacher 1 2 3 4 5 6 7 8 9 10 11 12 13