[leetcode] 819. Most Common Word @ python

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原题

Given a paragraph and a list of banned words, return the most frequent word that is not in the list of banned words. It is guaranteed there is at least one word that isn’t banned, and that the answer is unique.

Words in the list of banned words are given in lowercase, and free of punctuation. Words in the paragraph are not case sensitive. The answer is in lowercase.

Example:

Input:
paragraph = “Bob hit a ball, the hit BALL flew far after it was hit.”
banned = [“hit”]
Output: “ball”
Explanation:
“hit” occurs 3 times, but it is a banned word.
“ball” occurs twice (and no other word does), so it is the most frequent non-banned word in the paragraph.
Note that words in the paragraph are not case sensitive,
that punctuation is ignored (even if adjacent to words, such as “ball,”),
and that “hit” isn’t the answer even though it occurs more because it is banned.

Note:

1 <= paragraph.length <= 1000.
1 <= banned.length <= 100.
1 <= banned[i].length <= 10.
The answer is unique, and written in lowercase (even if its occurrences in paragraph may have uppercase symbols, and even if it is a proper noun.)
paragraph only consists of letters, spaces, or the punctuation symbols !?’,;.
There are no hyphens or hyphenated words.
Words only consist of letters, never apostrophes or other punctuation symbols.

解法

字符串处理, 将字符串先小写, 将所有标点符号都转化为空格, 然后将paragraph再转化为列表, 统计列表中不在banned的单词的频率, 找出频率最高的单词.
Time: O(n) + 2* O(m), n是字符串的长度, m是字典的长度
Space: O(1)

代码

class Solution(object):
    def mostCommonWord(self, paragraph, banned):
        """
        :type paragraph: str
        :type banned: List[str]
        :rtype: str
        """
        para = ''
        for char in paragraph.lower():
            if char in "!?',;.":
                para += ' '
            else:
                para += char
                
        count = collections.Counter([word for word in para.split() if word not in banned])
        common = max(count.values())
        for word in count:
            if count[word] == common:
                return word