Given a non-empty array of integers, return the k most frequent elements.

Example 1:

Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]

Example 2:

Input: nums = [1], k = 1
Output: [1]

Note:

You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
Your algorithm’s time complexity must be better than O(n log n), where n is the array’s size.
It’s guaranteed that the answer is unique, in other words the set of the top k frequent elements is unique.
You can return the answer in any order.

solution

time complexity : O(n)
space complexity : O(n)

class Solution {
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        // {num, freq}
        unordered_map<int, int> freqCnt;
        int maxFreq = 1;
        for (int i: nums)
            maxFreq = max(maxFreq, ++freqCnt[i]);
        
        // {freq, {nums}}
        unordered_map<int, vector<int>> freqNums;
        for (auto i: freqCnt)
            freqNums[i.second].push_back(i.first);
        
        vector<int> ans;
        for (int i=maxFreq; i>0; --i) {
            auto it = freqNums.find(i);
            if (it == freqNums.end()) continue;
            ans.insert(ans.end(), it->second.begin(), it->second.end());
            if (ans.size() == k) return ans;
        }
        
        return ans;
    }
};

流程：計算每個數字的頻率→根據每個頻率蒐集相對應的數字→頻率由大到小依序塞入答案中，直到k個。

因為可以假設k得出來的答案是獨一無二的，不會有這樣子的題目出現：

Input: nums = [1,2], k = 1
Output: [1]? [2]?
k must be 2, so that output = {1, 2}

所以給的k，一定是會把相同頻率數字全部蒐集起來。