819-最频繁的单词

Description:
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.

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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.

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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 !?’,;.
• Different words in paragraph are always separated by a space.
• There are no hyphens or hyphenated words.
• Words only consist of letters, never apostrophes or other punctuation symbols.

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问题描述

给定一个段落p和一个被禁止单词的列表，返回不在禁止列表中的最频繁的单词。至少有一个单词不在被禁止列表中，并且答案唯一

在被禁止列表中的单词为小写。段落中的单词大小写不敏感，返回结果需以小写形式

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问题分析

1. 去掉标点符号
2. 将段落转换为小写
3. 以空白符分隔段落
4. 通过HashMap计数

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解法

class Solution {
     public String mostCommonWord(String p, String[] banned) {
        Set<String> ban = new HashSet<>(Arrays.asList(banned));
        Map<String, Integer> count = new HashMap<>();
        // \\pP去除标点符号
        String[] words = p.replaceAll("\\pP" , "").toLowerCase().split("\\s+");
        String res = "";
        int max = 0;

        for (String w : words) {
            if (!ban.contains(w)) {
                count.put(w, count.getOrDefault(w, 0) + 1);
                if (count.get(w) > max) {
                    res = w;
                    max = count.get(w);
                }
            }
        }

        return res;
    }
}