題目描述：
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.
一刷題解（Dictionary）：
這題需要我們找出一數組裡出現次數最多的K個元素，因此，我們需要一個容器記錄「元素本身」和「元素出現頻次」兩個信息，那就非Dictionary莫屬了。首先遍歷數組，把所有元素本身為鍵，出現頻次為值記錄的字典裡，然後再讓字典根據值進行降序處理，那麼現在字典裡第一個元素肯定就是數組中出現頻次最多的元素。因此，我們再遍歷字典k次，依次把元素加到結果數組中即可。
public class Solution {
public int[] TopKFrequent(int[] nums, int k) {
Dictionary<int, int> freqDict = new Dictionary<int, int>();
//Dictionary for save freq of nums[i]
for (int i = 0; i < nums.Length; i++)
{
if (!freqDict.ContainsKey(nums[i]))
{
freqDict.Add(nums[i], 0);
}
freqDict[nums[i]]++;
}
//Sort Dictionary By Value(Descending)
var sortedFreqDict = freqDict.OrderByDescending(x => x.Value);
//Result array
int[] res = new int[k];
int counter = 0;
//Loop the sortedDictionary
foreach (var item in sortedFreqDict)
{
if(counter == k) { break; }
res[counter] = item.Key;
counter++;
}
return res;
}
}