[Leetcode] 819. Most Common Word 解题报告

这篇博客详细介绍了如何解决LeetCode中的819题,即找出一段文本中最频繁出现且不在禁用词汇列表中的单词。解题策略是利用哈希表记录单词出现次数,通过移除单词后的标点符号,确保正确计数。代码中使用了stringstream来高效地分割和处理单词。

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题目

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 !?',;.
  • 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.

思路

采用哈希表保存禁用词以及每个单词的出现个数即可。最后返回不出现在禁用词中,但是出现次数最多的单词。

可以采用多种方式对paragraph中的单词进行解析。我这里采用了stringtream,这样就可以按照空格将单词分割出来,非常方便。不过需要注意的是,我们需要去除某些单词后面附着的标点符号。

代码

class Solution {
public:
    string mostCommonWord(string paragraph, vector<string>& banned) {
        unordered_set<string> banned_words;         // for efficiency
        for (string &b : banned) {
            banned_words.insert(b);
        }
        unordered_map<string, int> hash;            // string -> appear count
        stringstream ss(paragraph);
        string s;
        while (ss >> s) {
            if (!isalpha(s.back())) {               // remove the punctuation
                s.pop_back();
            }
            for (int i = 0; i < s.length(); ++i) {  // convert to lower case
                s[i] = tolower(s[i]);
            }
            if (banned_words.count(s) == 0) {       // check whether it is banned
                ++hash[s];
            }
        }
        int max_count = 0;                          // find the max appearance
        string ret = "";
        for (auto it = hash.begin(); it != hash.end(); ++it) {
            if (it->second > max_count) {
                max_count = it->second;
                ret = it->first;
            }
        }
        return ret;
    }
};
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