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 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.

HashMapCount后 的思路类似：215. Kth Largest Element in an Array
http://www.jianshu.com/p/cbeeddb3cd1a

Solution1：HashMapCount + Bucketsort

思路： MapCount: num -> count，将count放到bucket中找出前k个num
Time Complexity: O(N) Space Complexity: O(N)

Solution2：HashMapCount + MaxHeap Sort

思路： MapCount: num -> count，建MaxHeap，将(num,count)放到堆中以count排出前k个num
Time Complexity: O(N + k * logN) N建堆
Space Complexity: O(N)

Solution3：HashMapCount + MinHeap 过滤维护前k个 最多count的元素

思路： MapCount: num -> count，建MaxHeap，将(num,count)放到堆中以count排出前k个num
Time Complexity: O(k + N * logk) k建堆
Space Complexity: O(N)

Solution4：HashMapCount + TreeMap作sort

思路： MapCount: num -> count，建TreeMap，积累 (count -> num_list)，因为TreeMap key=count有序 所以就可以选出最后最大的count的k个
Time Complexity: O(N * logN)
Space Complexity: O(N)
TreeMap概念可参考：http://www.jianshu.com/p/e29571903591

Solution1 Code：

class Solution {
    public List topKFrequent(int[] nums, int k) {
        
        // count
        Map frequencyMap = new HashMap();
        for (int n : nums) {
            frequencyMap.put(n, frequencyMap.getOrDefault(n, 0) + 1);
        }
        
        // bucket sort
        List[] bucket = new List[nums.length + 1];
        for (int key : frequencyMap.keySet()) {
            int frequency = frequencyMap.get(key);
            if (bucket[frequency] == null) {
                bucket[frequency] = new ArrayList<>();
            }
            bucket[frequency].add(key);
        }

        // prepare result
        List res = new ArrayList<>();

        outerloop:
        for (int pos = bucket.length - 1; pos >= 0; pos--) {
            if(bucket[pos] == null) continue;
            for(int i = 0; i < bucket[pos].size(); i++) {
                res.add(bucket[pos].get(i));
                if(res.size() == k) break outerloop;
            }
        }
        return res;
        
    }
}

Solution2 Code：

class Solution {
    public List topKFrequent(int[] nums, int k) {
        
        // count
        Map map = new HashMap();
        for (int n : nums) {
            map.put(n, map.getOrDefault(n, 0) + 1);
        }
        
        // max-heap to sort and get the Top K Frequent Elements
        PriorityQueue> maxHeap = 
                         new PriorityQueue<>((a,b)->(b.getValue()-a.getValue()));
        for(Map.Entry entry: map.entrySet()){
            maxHeap.add(entry);
        }
        
        List res = new ArrayList<>();
        while(res.size() < k){
            Map.Entry entry = maxHeap.poll();
            res.add(entry.getKey());
        }

        return res;
    }
}

Solution3 Code：

class Solution {
    public List topKFrequent(int[] nums, int k) {
        
        // count
        Map map = new HashMap();
        for (int n : nums) {
            map.put(n, map.getOrDefault(n, 0) + 1);
        }
        
        // min-heap to keep the Top K Frequent Elements
        PriorityQueue> minHeap = 
                         new PriorityQueue<>((a,b)->(a.getValue() - b.getValue()));
        for(Map.Entry entry: map.entrySet()) {
            minHeap.add(entry);
            if(minHeap.size() > k) {
                minHeap.poll();
            }
        }
        
        List res = new ArrayList<>();
        while(minHeap.size() > 0){
            Map.Entry entry = minHeap.poll();
            res.add(entry.getKey());
        }

        return res;
    }
}

Solution4 Code：

class Solution {
    public List topKFrequent(int[] nums, int k) {
        
        // count
        Map map = new HashMap();
        for (int n : nums) {
            map.put(n, map.getOrDefault(n, 0) + 1);
        }
        
        TreeMap> freqMap = new TreeMap<>();
        for(int num : map.keySet()){
           int freq = map.get(num);
           if(!freqMap.containsKey(freq)){
               freqMap.put(freq, new LinkedList<>());
           }
           freqMap.get(freq).add(num);
        }
        
        List res = new ArrayList<>();
        outerloop:
        while(true) { //since k is vaild
            Map.Entry> entry = freqMap.pollLastEntry();
            // if(entry == null) break;
            for(int i = 0; i < entry.getValue().size() ; i++) {
                res.add(entry.getValue().get(i));
                if(res.size() == k) break outerloop;
            }
        }
        return res;
    }
}