347. Top K Frequent Elements

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.

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;
    }
}