347. Top K Frequent Elements

My First PriorifyQueue Solution

This soltion use a Map to record the frequency of every number, and then use PriorityQueue to sort by frequency. The time complexity is O(nlogn), two pass.

    class NumCount{
        public NumCount(int num, int count){
            this.num = num;
            this.count = count;
        }
        public int num=0;
        public int count = 0;
    }
    public int[] topKFrequent(int[] nums, int k) {
        Map<Integer, Integer> map = new HashMap<>();
        for(int i=0;i<nums.length;i++){
            map.putIfAbsent(nums[i], 0);
            map.put(nums[i], map.get(nums[i])+1);
        }

        PriorityQueue<NumCount> queue = new PriorityQueue<>((a, b)->b.count-a.count);
        Set<Integer> set = map.keySet();
        for(int num: set){
            NumCount n = new NumCount(num, map.get(num));
            queue.add(n);
        }

        int[] res = new int[k];
        for(int i=0;i<k;i++){
            res[i]=queue.poll().num;
        }
        return res;
    }

My Second Bucket Sort Solution

The time conplexity is O(n), because the sorting has been done when put the numbers in buckets.

    public int[] topKFrequent(int[] nums, int k) {
        List<Integer>[] bucket = new List[nums.length+1];
        
        Map<Integer, Integer> map = new HashMap<>();
        for(int i=0;i<nums.length;i++){
            map.putIfAbsent(nums[i], 0);
            map.put(nums[i], map.get(nums[i])+1);
        }
        
        Set<Integer> set = map.keySet();
        for(int num: set){
            int freq = map.get(num);
            if(bucket[freq]==null){
                bucket[freq]=new ArrayList<>();
            }
            bucket[freq].add(num);
        }
        int[] res = new int[k];
        
        int j=0;
        for(int i=bucket.length-1;i>=0;i--){
            if(bucket[i]==null)
                continue;
            List<Integer> list = bucket[i];
            for(int n: list){
                if(j>=k)
                    return res;
                res[j++]=n;
            }
        }
        return res;
    }

 Or a Queue can be used:

class Solution {
    public int[] topKFrequent(int[] nums, int k) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            map.put(nums[i], map.getOrDefault(nums[i], 0) + 1);
        }

        Queue<Integer>[] buckets = new LinkedList[nums.length + 1];
        Set<Integer> set = map.keySet();
        for (int n : set) {
            int freq = map.get(n);
            if (buckets[freq] == null) {
                buckets[freq] = new LinkedList<>();
            }
            buckets[freq].add(n);
        }
        int[] res = new int[k];
        int j = 0;
        for (int i = buckets.length - 1; i >= 0; i--) {
            if (buckets[i] == null)
                continue;
            Queue<Integer> queue = buckets[i];
            while (!queue.isEmpty()) {
                if (j >= k)
                    return res;
                res[j++] = queue.poll();
            }

        }
        return res;
    }
}