LintCode 1484: The Most Frequent Word (练习C++ string, map和set)

原题如下：
1484. The Most Frequent word
Give a string s witch represents the content of the novel, and then give a list indicating that the words do not participate in statistics, find the most frequent word(return the one with the smallest lexicographic order if there are more than one word)

Example
Input: s = “Jimmy has an apple, it is on the table, he like it”
excludeWords = [“a”,“an”,“the”]
Output:“it”

我的解法如下：
注意：

1. string.find(substring, pos) 返回-1 (即string::npos)如果没有找到substring。
2. 注意要跳过’ ', ',‘和’.'字符。
3. 有三种情况需要区分:
   a. maxCountWord还没有初始化
   b. 新找到的词的词频已经大于maxCount
   c. 新找到的词的词频等于maxCount，则要比较该词与maxCountWord的大小。

class Solution {
public:/*** @param s: a string* @param excludewords: a dict* @return: the most frequent word*/string frequentWord(string &s, unordered_set &excludewords) {map m; //word vs countint len = s.size();int maxCount = 0;string maxCountWord = "";int i = 0, j = 0;string word;while (i < len) {if (s[i] == ' ') while(s[i++] == ' ');j = s.find(' ', i);if (j != string::npos) {   //if found the ' 'if ((s[j - 1] == ',') || (s[j - 1] == '.')) word = s.substr(i, j - i - 1);else word = s.substr(i, j - i);} else { //it should be the last wordif ((s[len - 1] == ',') || (s[len - 1] == '.')) word = s.substr(i, len - i - 1);else word = s.substr(i, len - i);}if (excludewords.find(word) == excludewords.end()) {if (m.find(word) == m.end()) {m[word] = 1;} else {m[word]++;}if (maxCountWord.size() == 0) {maxCount = m[word];maxCountWord = word;} else if ((m[word] > maxCount)) {maxCount = m[word];maxCountWord = word;} else if (m[word] == maxCount) {if (word < maxCountWord)maxCountWord = word;}}if (j == string::npos) return maxCountWord;   //j = -1, reach the end of the stringif (j > 0) i = j + 1;else i++;}return maxCountWord;}
};

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