LeetCode Most Common Word 最常见的词

高频非禁用词查找

最新推荐文章于 2025-09-12 17:31:31 发布

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原文链接：https://segmentfault.com/a/1190000014392925

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#数据结构与算法

本文介绍了一种算法，用于从给定的段落中找出出现频率最高的非禁用词。该算法首先清理输入文本中的标点符号并转换为小写形式，接着统计每个单词的出现次数，并排除掉禁用词列表中的单词。

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 !?',;.
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.

这是一道字符串处理的题目，给我们一个字符串段落，以及一个被禁止的单词，段落中我们要处理!?',;.这些标点符号。我们可以算出各个单词出现的次数，然后将被禁止的单词排除在外，就可以算出出现次数最多的单词的哪一个。
解法一：

class Solution {
public:
    string mostCommonWord(string paragraph, vector<string>& banned) {
        map<string,int> M;
        map<string,int>::iterator it;
        M.clear();
        int i,j,n=paragraph.size(),m=banned.size();
        string s;
        for(i=0;i<paragraph.length();i++)if(paragraph[i]>='A'&&paragraph[i]<='Z')paragraph[i]+='a'-'A';
        for(i=0;i<n;i=j)
        {
            s="";
            for(j=i;j<n&&(paragraph[j]<'a'||paragraph[j]>'z');j++);
            if(j==n)break;
            for(;j<n&&paragraph[j]>='a'&&paragraph[j]<='z';j++)s+=paragraph[j];
            M[s]++;
        }
        for(i=0;i<m;i++)M[banned[i]]=0;
        for(it=M.begin(),i=0;it!=M.end();it++)if(it->second>i)
        {
            s=it->first;
            i=it->second;
        }
        return s;
    }
};