可以暴力解决，但是为了锻炼一下ac自动机的编程，我们使用ac自动机。
ac自动机主要维护两个列表，一个列表ch，ch[f][idx]表示从父节点f向idx这个方向走，走到的节点。另一个列表nex，nex[i]表示节点i回跳边的节点。
from collections import defaultdict, deque
class Solution:def addBoldTag(self, s: str, words: List[str]) -> str:p = ''.join(words)n = len(p)n1 = len(s)r1 = []# ac自动机ch = defaultdict(dict)nex = [0]*(n+1)cnt = [0]*(n+1)idy = 0def assign(s):if s.isdigit():return int(s)elif ord('a') <= ord(s) <= ord('z'):return ord(s) - ord('a') + 10else:return ord(s) - ord('A') + 36def insert(word):f = 0nonlocal idyfor s in word:idx = assign(s)if idx not in ch[f]:idy += 1ch[f][idx] = idyf = idyelse:f = ch[f][idx]cnt[f] = len(word)def bulid():qu = deque()for i in range(62):if i not in ch[0]:continuequ.append(ch[0][i])while qu:f = qu.popleft()for i in range(62):if i not in ch[f]:ch[f][i] = 0 if i not in ch[nex[f]] else ch[nex[f]][i]else:nex[ch[f][i]] = 0 if i not in ch[nex[f]] else ch[nex[f]][i]qu.append(ch[f][i])def query(s):f = 0for i in range(n1):idx = assign(s[i])f = 0 if idx not in ch[f] else ch[f][idx]idy = fwhile idy != 0:if cnt[idy]:r1.append([i-cnt[idy]+1, i+1])breakidy = nex[idy]return for word in words:insert(word)bulid()query(s)r1.sort()leth = len(r1)if leth == 0:return srec = [r1[0]]for idx in range(1, leth):l, r = r1[idx]if l <= rec[-1][1]:rec[-1][1] = max(r, rec[-1][1])else:rec.append([l,r])dic = defaultdict(str)for l, r in rec:dic[l] = '<b>'dic[r] = '</b>'ans = ''for idx in range(n1):if dic[idx]:ans += dic[idx]ans += s[idx]if dic[n1]:ans += dic[n1]return ans确实是通过了，但是！！！暴力解法居然比ac自动机更快！！！哪边出了问题？？？
下面是暴力的，上面是ac自动机
暴力代码：
class Solution:def addBoldTag(self, s: str, words: List[str]) -> str:from collections import defaultdictr1 = []l1 = len(s)for word in words:l2 = len(word)for idx in range(l1-l2+1):if s[idx:idx+l2] == word:r1.append([idx,idx+l2])r1.sort()leth = len(r1)if leth == 0:return srec = [r1[0]]for idx in range(1, leth):l, r = r1[idx]if l <= rec[-1][1]:rec[-1][1] = max(r, rec[-1][1])else:rec.append([l,r])dic = defaultdict(str)for l, r in rec:dic[l] = '<b>'dic[r] = '</b>'ans = ''for idx in range(l1):if dic[idx]:ans += dic[idx]ans += s[idx]if dic[l1]:ans += dic[l1]return ans
