819. Most Common Word - Easy-优快云博客

本文介绍了一种算法，用于从给定段落中找出最常出现且未被列入禁用词列表的单词。通过使用正则表达式进行单词分割，并利用HashMap统计单词频率，最终在排除所有禁用词后，找到出现次数最多的单词。

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.

用split + 正则表达式把paragraph的单词分离再转成小写，用hashmap保存paragraph中的单词，遍历banned，在hashmap中去除相应key，最后找出hashmap中出现次数最多的key

注意正则表达式的用法：

[xyz]　　字符集，匹配包含的任一字符

+　　　一次或多次匹配前面的字符或子表达式

time: O(P + B), space: O(P + B)　　-- P: length of paragraph, B: length of banned

class Solution {
    public String mostCommonWord(String paragraph, String[] banned) {
        String res = "";
        String[] words = paragraph.toLowerCase().split("[ !?',;.]+");
        HashMap<String, Integer> map = new HashMap<>();
        for(String word : words) {
            map.put(word, map.getOrDefault(word, 0) + 1);
        }
        for(String word : banned) {
            if(map.containsKey(word))
                map.remove(word);
        }
        
        for(String word : map.keySet()) {
            System.out.println(word);
            if(res == "" || map.get(word) > map.get(res))
                res = word;
        }
        return res;
    }
}