347. Top K Frequent Elements

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

For example,
Given [1,1,1,2,2,3] and k = 2, return [1,2] .

Note:

• You may assume K are 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.

Using Count Sort

 Publicilist<int> Topkfrequent (int[] Nums,intk) {varres =Newlist<int>(); if(Nums. Count () = =0)returnRes; varHashtable =Newdictionary<int,int>();  for(inti =0;i< Nums. Count (); i++)        {            if(Hashtable. ContainsKey (Nums[i]) hashtable[nums[i]]++; ElseHashtable. ADD (Nums[i],1); }        //Sort by Count sort        varDP =Newlist<int>(); foreach(varPairinchHashtable) {DP. ADD (pair.        Value); }        intA =FINDK (dp,k); foreach(varPairinchHashtable) {            if(pair.) Value >=a) Res. ADD (pair.        Key); }        returnRes; }         Public intFINDK (list<int> Nums,intk) {if(Nums. Count () = =0)return 0; intmax = nums[0]; intMin = nums[0]; varb =New int[Nums.        Count ()]; foreach(intNinchnums) {Max=Math.max (max,n); Min=math.min (min,n); }                intKK = Max-min +1; varc =New int[KK];  for(inti =0; I<nums. Count (); i++) {C[nums[i]-min] + =1; }         for(inti =1; I<c.count (); i++) {C[i]= C[i] + c[i-1]; }         for(inti = Nums. Count ()-1; I >=0; i--) {b[--c[nums[i]-min]] =Nums[i]; }        returnB[nums. Count ()-K]; }

347. Top K Frequent Elements