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Here is an optimized version of the thought process and solution:

Final Solution Code

import java.util.HashMap;import java.util.HashSet;import java.util.Map;public class Solution {    public String mostCommonWord(String paragraph, String[] banned) {        // Convert paragraph to lowercase and remove punctuation        StringBuilder cleanParagraph = new StringBuilder();        for (char c : paragraph.toCharArray()) {            if (c >= 'a' && c <= 'z') {                cleanParagraph.append(c);            }        }        // Split into words        String[] words = cleanParagraph.toString().split(" +");        // Count frequency of each word        Map
   
     frequencyMap = new HashMap<>();        for (String word : words) {            frequencyMap.put(word, frequencyMap.getOrDefault(word, 0) + 1);        }        // Create banned words set for quick lookup        HashSet
    
      bannedWords = new HashSet<>();        for (String bw : banned) {            bannedWords.add(bw.toLowerCase());        }        // Exclude banned words and find the most frequent        int maxCount = -1;        String result = "";        for (Map.Entry
     
       entry : frequencyMap.entrySet()) {            if (!bannedWords.contains(entry.getKey())) {                if (entry.getValue() > maxCount) {                    maxCount = entry.getValue();                    result = entry.getKey();                }            }        }        return result;    }}
     
    
   

Explanation

• The code first processes the input paragraph to remove punctuation and convert it to lowercase, ensuring uniformity in word processing.
• It then splits the cleaned string into individual words and uses a hash map to count each word's occurrences.
• Banned words are stored in a set for quick exclusion.
• Finally, the code iterates through the frequency map, excluding banned words, and identifies the word with the highest count, which is then returned as the result.

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