LeetCode Most Common Word 最常见的词-白红宇

LeetCode Most Common Word 最常见的词

阅读量：801 次

发布时间：2023-01-31

本文共 3035 字，大约阅读时间需要 10 分钟。

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

Data Preparation

Convert the entire paragraph to lowercase to handle case insensitivity.

Remove all punctuation marks (such as commas, periods, exclamation points, etc.) to isolate words.

Ensure words are properly separated by spaces to avoid partial words (e.g., "ball," becomes "ball").

Word Frequency Calculation

Traverse the prepared string, extracting each word by ignoring punctuation and case differences.

Use a hash map (dictionary) to count occurrences of each word.

For each character in the paragraph: If it's a letter, add it to the current word being built. If it's not a letter or reaches the end of the string, finalize the word and update its count in the hash map.

Filter Banned Words

Store banned words in a set for quick lookup.

Iterate through the hash map to exclude any words that exist in the banned set, keeping only valid words.

Determine Most Frequent Word

Sort the remaining words by their frequency in descending order.

Return the first word in this sorted list, as it by definition is unique and has the highest count according to the problem constraints.

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.

转载地址：http://oogyk.baihongyu.com/

你可能感兴趣的文章