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 = 0
        def assign(s):
            if s.isdigit():
                return int(s)
            elif ord('a') <= ord(s) <= ord('z'):
                return ord(s) - ord('a') + 10
            else:
                return ord(s) - ord('A') + 36

        def insert(word):
            f = 0
            nonlocal idy
            for s in word:
                idx = assign(s)
                if idx not in ch[f]:
                    idy += 1
                    ch[f][idx] = idy
                    f = idy
                else:
                    f = ch[f][idx]
            cnt[f] = len(word)

        def bulid():
            qu = deque()
            for i in range(62):
                if i not in ch[0]:
                    continue
                qu.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 = 0
            for i in range(n1):
                idx = assign(s[i])
                f = 0 if idx not in ch[f] else ch[f][idx]
                idy = f
                while idy != 0:
                    if cnt[idy]:
                        r1.append([i-cnt[idy]+1, i+1])
                        break
                    idy = nex[idy]
            return 


        for word in words:
            insert(word)
        bulid()
        query(s)
        r1.sort()
        leth = len(r1)
        if leth == 0:return s
        rec = [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 defaultdict
        r1 = []
        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 s
        rec = [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