leetcode 819. Most Common Word

Posted 2021-02-08 jamieliu

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.

 

解法一：正则表达式+拆成一个一个word

4 steps:

1. remove all punctuations
2. change to lowercase
3. words count for each word not in banned set
4. return the most common word

学到了很多java里的新方法。

还用到了正则表达式    W+表示获取一个单词后面的一个字符  s+表示空白字符（包括空格键 制表符等）

这个代码很简洁，但是很慢。

class Solution {
    public String mostCommonWord(String paragraph, String[] banned) {
        Set<String> ban = new HashSet<>(Arrays.asList(banned));
        Map<String,Integer> count = new HashMap<>();
        String[] words = paragraph.replaceAll("\W+", " ").toLowerCase().split("\s+");
        for(String w: words) if(!ban.contains(w)) count.put(w, count.getOrDefault(w,0)+1);
        return Collections.max(count.entrySet(), Map.Entry.comparingByValue()).getKey();
    }
}

 

解法二：拆成一个一个char + stringBuilder()

class Solution {
    public String mostCommonWord(String paragraph, String[] banned) {
        paragraph += ".";

        Set<String> banset = new HashSet();
        for (String word: banned) banset.add(word);
        Map<String, Integer> count = new HashMap();

        String ans = "";
        int ansfreq = 0;

        StringBuilder word = new StringBuilder();
        for (char c: paragraph.toCharArray()) {
            if (Character.isLetter(c)) {
                word.append(Character.toLowerCase(c));
            } else if (word.length() > 0) {
                String finalword = word.toString();
                if (!banset.contains(finalword)) {
                    count.put(finalword, count.getOrDefault(finalword, 0) + 1);
                    if (count.get(finalword) > ansfreq) {
                        ans = finalword;
                        ansfreq = count.get(finalword);
                    }
                }
                word = new StringBuilder();
            }
        }

        return ans;
    }
}