Top K Frequent Elements
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.
It's guaranteed that the answer is unique, in other words the set of the top k frequent elements is unique.
You can return the answer in any order.
class Solution {
public int[] topKFrequent(int[] nums, int k) {
Map<Integer, Integer> map = new HashMap<>();
for (int x : nums)
map.put(x, map.getOrDefault(x, 0) + 1);
ArrayList<Integer>[] list = new ArrayList[nums.length + 1];
for (int x : map.keySet()) {
if (list[map.get(x)] == null)
list[map.get(x)] = new ArrayList<>();
list[map.get(x)].add(x);
}
int[] ans = new int[k];
int index = 0;
for (int i = list.length - 1; i >= 0 && k > 0; i--) {
if (list[i] != null) {
for (int j = 0; k > 0 && j < list[i].size(); j++) {
ans[index++] = list[i].get(j);
k--;
}
if (k == 0)
break;
}
}
return ans;
}
}
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