题目
给定一个段落 (paragraph) 和一个禁用单词列表 (banned)。返回出现次数最多,同时不在禁用列表中的单词。
题目保证至少有一个词不在禁用列表中,而且答案唯一。
禁用列表中的单词用小写字母表示,不含标点符号。段落中的单词不区分大小写。答案都是小写字母。
示例:
输入: paragraph = "Bob hit a ball, the hit BALL flew far after it was hit."
banned = ["hit"]
输出: "ball"
解释: "hit"
出现了3次,但它是一个禁用的单词。"ball"
出现了2次 (同时没有其他单词出现2次),所以它是段落里出现次数最多的,且不在禁用列表中的单词。
注意,所有这些单词在段落里不区分大小写,标点符号需要忽略(即使是紧挨着单词也忽略, 比如 "ball,"
), "hit"
不是最终的答案,虽然它出现次数更多,但它在禁用单词列表中。
提示:
1 <= 段落长度 <= 1000
0 <= 禁用单词个数 <= 100
1 <= 禁用单词长度 <= 10
- 答案是唯一的, 且都是小写字母 (即使在
paragraph
里是大写的,即使是一些特定的名词,答案都是小写的。) paragraph
只包含字母、空格和下列标点符号!?',;.
- 不存在没有连字符或者带有连字符的单词。
-
方案一(Counter)
class Solution:
def mostCommonWord(self, paragraph: str, banned: List[str]) -> str:
c = Counter()
word = []
for ch in paragraph:
ch = ch.lower()
if ord("a") <= ord(ch) <= ord("z"):
word.append(ch)
elif word:
c["".join(word)] += 1
word = []
if word:
c["".join(word)] += 1
banned_set = set(banned)
for word, _ in c.most_common():
if word not in banned_set:
return word
return ""
或
class Solution(object):
def mostCommonWord(self, paragraph, banned):
banset = set(banned)
for c in "!?',;.":
paragraph = paragraph.replace(c, " ")
count = collections.Counter(
word for word in paragraph.lower().split())
ans, best = '', 0
for word in count:
if count[word] > best and word not in banset:
ans, best = word, count[word]
return ans
参考链接
https://leetcode-cn.com/explore/featured/card/2020-top-interview-questions/281/string/1259/