[leetcode]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

分析思路：

借鉴统计字符串中每个字符的个数的思想，用频数作为数组的下标

1.利用map结构统计每个数字的频数

2.初始化list数组，注意这里数组长度是输入数字的个数

3.以频数作为数组下标，初始化list，并添加相应的key

4.倒序添加频数多的，直到个数大于k结束

public class Solution {    public List<Integer> topKFrequent(int[] nums, int k) {        if(nums==null||nums.length==0||k<=0){return null;}Map<Integer,Integer> frequencyMap = new HashMap<Integer,Integer>();/**************get each element frequency *****************/for(int n:nums){if(frequencyMap.containsKey(n)){frequencyMap.put(n, frequencyMap.get(n)+1);}elsefrequencyMap.put(n, 1);}List<Integer>[] bucket = new List[nums.length];/******************generate list[] by frequency*****************/for(int key:frequencyMap.keySet()){int v = frequencyMap.get(key);if(bucket[v-1]==null){bucket[v-1] = new ArrayList<Integer>();}bucket[v-1].add(key);}List<Integer> result = new ArrayList<Integer>();label:for(int i=bucket.length-1;i>=0;i--){if(bucket[i]!=null){for(Integer num:bucket[i]){if(result.size()<k){result.add(num);}elsebreak label;}}}return result;    }}