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.
思路：先将所有单词放入map中（已根据字母表递增排好序），统计出现的次数，开启n个桶，第i个桶存放出现次数为i的单词，然后从最后一个桶开始拿出其中的单词插入结果数组的尾部。
class Solution {
public:
vector<string> topKFrequent(vector<string>& words, int k) {
map<string,int> cnt;
for(auto& i:words) cnt[i]++; //&
vector<vector<string> >bucket(words.size());
for(auto i:cnt) bucket[i.second].push_back(i.first);
vector<string> res;
for(int i=(int)words.size()-1;i>=0;i--){
int n = min(k,(int)bucket[i].size());
res.insert(res.end(),bucket[i].begin(),bucket[i].begin()+n);
k -= n;
}
return res;
}
};