leetcode 347. Top K Frequent Elements

题目内容

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.

题目分析

通过hashmap确定所有值出现的频率，通过频率的排序来找出top k。建立数组，以数组的index作为频率的值。

public class Solution {
public List<Integer> topKFrequent(int[] nums, int k) {
List<Integer>[] buk = new List[nums.length+1];//注意这里的+1
Map<Integer,Integer> frequencyMap = new HashMap<Integer,Integer>();//构建hashmap
for(int n : nums)
frequencyMap.put(n,frequencyMap.getOrDefault(n,0)+1);//找到频率

for(int key : frequencyMap.keySet()){
int frequency = frequencyMap.get(key);

if(buk[frequency]==null)
buk[frequency]=new ArrayList<Integer>();//以频率的值为index初始化list
buk[frequency].add(key);
}
List<Integer> ans =new ArrayList<Integer>();
for(int pos=buk.length-1;pos>=0&&ans.size()<k;pos--)
{
if(buk[pos]!=null) ans.addAll(buk[pos]);
}

return ans;
}
}