Leetcode: 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.

很像majority element III, 但是那道题有O(k) extra space的限制，这里没有。有任意extra space, 同时知道elem range情况下，bucket sort最节省时间

语法上注意第5行，建立一个ArrayList的array，后面没有泛型generic

List[] bucket = new List[nums.length + 1];  or

List<Integer>[] = new List[nums.length + 1];

Time Complexity： O（N）

class Solution {
    public List<Integer> topKFrequent(int[] nums, int k) {
        List<Integer> res = new ArrayList<>();
        if (nums == null || nums.length == 0) return res;
        
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        for (int num : nums) {
            if (map.containsKey(num)) {
                map.put(num, map.get(num) + 1);
            }
            else map.put(num, 1);
        }
        
        List[] bucket = new List[nums.length + 1];
        
        for (int key : map.keySet()) {
            int freq = map.get(key);
            if (bucket[freq] == null) {
                bucket[freq] = new ArrayList<Integer>();
            }
            bucket[freq].add(key);
        }
        
        for (int i = bucket.length - 1; i >= 0; i --) {
            if (bucket[i] == null) continue;
            if (bucket[i].size() + res.size() > k) break;
            res.addAll(bucket[i]);
        }
        return res;
    }
}

方法2： MaxHeap, 用map.entrySet(), return Map.Entry type, 可以调用getKey() 和 getValue()

Time Complexity: O(NlogN)

public class Solution {
    public List<Integer> topKFrequent(int[] nums, int k) {
        List<Integer> res = new ArrayList<Integer>();
        Map<Integer, Integer> freqMap = new HashMap<Integer, Integer>();
        
        for (int elem : nums) {
            if (freqMap.containsKey(elem)) {
                freqMap.put(elem, freqMap.get(elem)+1);
            }
            else {
                freqMap.put(elem, 1);
            }
        }

        Comparator<Map.Entry<Integer, Integer>> comp = new Comparator<Map.Entry<Integer, Integer>>() {
            public int compare(Map.Entry<Integer, Integer> entry1, Map.Entry<Integer, Integer> entry2) {
                return entry2.getValue() - entry1.getValue();
            }
        };
        
        PriorityQueue<Map.Entry<Integer, Integer>> maxHeap = new PriorityQueue<Map.Entry<Integer, Integer>>(11, comp);
        
        for (Map.Entry<Integer, Integer> each : freqMap.entrySet()) {
            maxHeap.offer(each);
        }
        
        for (int i=1; i<=k; i++) {
            res.add(maxHeap.poll().getKey());
        }
        return res;
    }
}