[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);
}else
frequencyMap.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);
}else
break label;

}
}
}
return result;
}
}