819. Most Common Word

问题描述

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.

题目链接：

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思路分析

找到一个段落当中的，在除了被ban掉的词之外，出现次数最多的词。

使用map和isalpha()配合，统计每个词出现的次数，然后再将ban掉的词的次数清零，最后循环找到map中出现次数最多的词。

代码

class Solution {
public:
    string mostCommonWord(string paragraph, vector<string>& banned) {
        unordered_map<string, int> map;
        for (int i = 0; i < paragraph.size();){
            string word = "";
            while(i < paragraph.size() && isalpha(paragraph[i])) word.push_back(tolower(paragraph[i++]));
            while(i < paragraph.size() && !isalpha(paragraph[i])) i++;
            map[word]++;
        }

        for (auto word : banned){
            map[word] = 0;
        }

        string result = "";
        int count = 0;
        for (auto word : map){
            if (word.second > count){
                count = word.second;
                result = word.first;
            }
        }

        return result;
    }
};

时间复杂度：O(n)
空间复杂度：O(n)

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反思

学到了isalpah()的用法，借助一些c++自己库中的功能，可以很轻松的完成算法。