347. Top K Frequent Elements - Medium

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.

 

M1: use maxHeap. Put entry into maxHeap so we can always poll a number with the largest frequency

time: O(nlogn) 　　-- O(n) for traverse, O(logn) for adding elements in heap, space: O(n)

class Solution {
    public List<Integer> topKFrequent(int[] nums, int k) {
        HashMap<Integer, Integer> map = new HashMap<>();
        for(int n : nums) {
            map.put(n, map.getOrDefault(n, 0) + 1);
        }
        
        PriorityQueue<Map.Entry<Integer, Integer>> maxHeap = new PriorityQueue<>((a, b) -> b.getValue() - a.getValue());
        for(Map.Entry<Integer, Integer> entry: map.entrySet()) {
            maxHeap.add(entry);
        }
        
        List<Integer> res = new ArrayList<>();
        while(res.size() < k) {
            res.add(maxHeap.poll().getKey());
        }
        return res;
    }
}

 

M2: use treeMap documents

treemap基于红黑树，检索时间是O(logn)，默认按照key的自然大小排序（升序，从小到大）

Use frequency as the key so we can get all frequencies in order，先用hashmap统计频率，再放入treemap <freq, n>，最后取出前k个value即可

time: O(nlogn), space: O(n)

class Solution {
    public List<Integer> topKFrequent(int[] nums, int k) {
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        for(int n : nums) {
            map.put(n, map.getOrDefault(n, 0) + 1);
        }
        
        TreeMap<Integer, List<Integer>> treemap = new TreeMap<>();
        for(Map.Entry<Integer, Integer> entry : map.entrySet()) {
            int num = entry.getKey(), freq = entry.getValue();
            if(!treemap.containsKey(freq)) {
                treemap.put(freq, new ArrayList<>());
            }
            treemap.get(freq).add(num);
        }
        
        List<Integer> res = new ArrayList<>();
        while(res.size() < k) {
            List<Integer> list = treemap.pollLastEntry().getValue();
            for(int i = 0; i < list.size(); i++) {
                if(res.size() < k)
                    res.add(list.get(i));
                else
                    break;
            }
        }
        return res;
    }
}