LeetCode 692. Top K Frequent Words

https://leetcode.com/problems/top-k-frequent-words/description/

---

Given a non-empty list of words, return the k most frequent elements.

Your answer should be sorted by frequency from highest to lowest. If two words have the same frequency, then the word with the lower alphabetical order comes first.

Example 1:

Input: ["i", "love", "leetcode", "i", "love", "coding"], k = 2
Output: ["i", "love"]
Explanation: "i" and "love" are the two most frequent words.
    Note that "i" comes before "love" due to a lower alphabetical order.

Example 2:

Input: ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"], k = 4
Output: ["the", "is", "sunny", "day"]
Explanation: "the", "is", "sunny" and "day" are the four most frequent words,
    with the number of occurrence being 4, 3, 2 and 1 respectively.

Note:
You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
Input words contain only lowercase letters.
Follow up:
Try to solve it in O(n log k) time and O(n) extra space.

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寻找最频繁出现的k个字符串。
创建一个map< string, int>对象，将vector中的字符串存入map中。然后每次存入前都查找当前字符串，第一次存进去对应的value设置为1，若已存在则在原来value上加1.
对map进行排序：
将map里的值再次存入一个vector < pair< string, int>>对象中，再对其用sort函数排序。构造一个比较函数cmpByValue()，按值大小排序。
struct cmpByValue {
bool operator()(const pair& a, const pair& b) {
if (a.second > b.second) return true;
if (a.second == b.second) return a.first < b.first;
return false;
}
};
再返回排序好的前k个字符串。

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class Solution {
public:
    struct cmpByValue {  
        bool operator()(const pair<string, int>& a, const pair<string, int>& b) {
            if (a.second > b.second) return true;
            if (a.second == b.second) return a.first < b.first;
            return false;
        }
    };
    vector<string> topKFrequent(vector<string>& words, int k) {
        map<string, int> m;
        for (int i = 0; i < words.size(); i++) {
            m[words[i]]++;
        }
        vectorstring, int> > v(m.begin(), m.end());
        vector<string> s;
        sort(v.begin(), v.end(), cmpByValue());
        for (int i = 0; i < k; i++) s.push_back(v[i].first);
        return s;
    }
};

110 / 110 test cases passed.
Status: Accepted
Runtime: 16 ms