LeetCode题解(0819)：词频分析(Python)

题目：原题链接（简单）

标签：字符串、哈希表

解法时间复杂度空间复杂度执行用时Ans 1 (Python)$O(N+B)$ : N为句子长度、B为禁用列表长度$O(N+B)$40ms (92.22%)Ans 2 (Python)Ans 3 (Python)

解法一（哈希表）：

def mostCommonWord(self, paragraph: str, banned: List[str]) -> str: hashmap = {} for word in re.split("[ ,.?!]", paragraph.lower()): word = "".join(list(filter(str.isalpha, word))) if len(word) > 0: if word not in hashmap: hashmap[word] = 1 else: hashmap[word] += 1 times = 0 ans = "" for key, value in hashmap.items(): if value > times and key not in banned: ans = key times = value return ans