README_EN.md

347. Top K Frequent Elements

中文文档

Description

Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.

 

Example 1:

Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]

Example 2:

Input: nums = [1], k = 1
Output: [1]

 

Constraints:

• 1 <= nums.length <= 105
• k is in the range [1, the number of unique elements in the array].
• It is guaranteed that the answer is unique.

 

Follow up: Your algorithm's time complexity must be better than O(n log n), where n is the array's size.

Solutions

Python3

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        counter = collections.Counter(nums)
        buckets = [[] for _ in range(len(nums) + 1)]
        for num, freq in counter.items():
            buckets[freq].append(num)
        i, res = len(nums), []
        while k > 0 and i >= 0:
            if buckets[i]:
                for num in buckets[i]:
                    if k <= 0:
                        break
                    res.append(num)
                    k -= 1
            i -= 1
        return res

Java

class Solution {
    public int[] topKFrequent(int[] nums, int k) {
        Map<Integer, Integer> counter = new HashMap<>();
        for (int num : nums) {
            counter.put(num, counter.getOrDefault(num, 0) + 1);
        }
        List<Integer>[] buckets = new List[nums.length + 1];
        for (Map.Entry<Integer, Integer> entry : counter.entrySet()) {
            int num = entry.getKey();
            int freq = entry.getValue();
            if (buckets[freq] == null) {
                buckets[freq] = new ArrayList<>();
            }
            buckets[freq].add(num);
        }
        int[] res = new int[k];
        for (int i = nums.length; i >= 0 && k > 0; --i) {
            if (buckets[i] != null) {
                for (int num : buckets[i]) {
                    if (k <= 0) {
                        break;
                    }
                    res[--k] = num;
                }
            }
        }
        return res;
    }
}

...