347. Top K Frequent Elements

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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.

这道题用Bucket Sort，新建一个List<Integer>[nums.length + 1]，再新建一个hashmap把所有的数和出现的次数存下来，再以出现次数为索引把把数存到相应的List<Integer>[i]里，在从后向前遍历k个List<Integer>输出。代码如下：

public class Solution {    public List<Integer> topKFrequent(int[] nums, int k) {        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();        List<Integer>[] list = new List[nums.length + 1];        List<Integer> res = new ArrayList<Integer>();        for (int num: nums) {            map.put(num, map.getOrDefault(num, 0) + 1);        }        for (int num: map.keySet()) {            if (list[map.get(num)] == null) {                list[map.get(num)] = new ArrayList<Integer>();            }            list[map.get(num)].add(num);        }        for (int i = nums.length; res.size() < k && i >= 0; i --) {            if (list[i] != null) {                res.addAll(list[i]);            }        }        return res;    }}

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