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 logn), where n is the array's size.

思路：通过可保存映射关系的map，将每个值及其出现次数保存下来，然后就是对entry排序的问题，在这道题中是按值排序。

为了使用Collection的sort方法，先将entrySet转化为List，然后重写了Comparator接口的compare方法，取降序排序后的前k项就是我们要的结果。

public class Solution {     public List<Integer> topKFrequent(int[] nums, int k) {        Map<Integer,Integer> count = new HashMap<Integer,Integer>();        for(int i=0;i<nums.length;i++){            if(count.containsKey(nums[i])) count.put(nums[i], count.get(nums[i])+1);            else count.put(nums[i],1);        }                List<Map.Entry<Integer, Integer>> list = new ArrayList<Map.Entry<Integer,Integer>>(count.entrySet());        Collections.sort(list, new Comparator<Map.Entry<Integer, Integer>>() {              //降序排序              @Override              public int compare(Map.Entry<Integer, Integer> o1, Map.Entry<Integer, Integer> o2) {                   return o2.getValue().compareTo(o1.getValue());              }          });          List<Integer> result = new ArrayList<Integer>();for(int i=0;i<k;i++){result.add(list.get(i).getKey());}return result;    }}

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