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：

定义一个类 保存一个num的值和频率，并实现比较方法

遍历数组，并用hashMap保存各个数字出现的频率

对hashmap的value集合排序，获取频率最高的k位

public List<Integer> topKFrequent(int[] nums, int k) {
List<Integer> res = new ArrayList<Integer>();
Map<Integer,FEntity> map = new HashMap<Integer,FEntity>();
//统计各数字出现的次数 存入hashmap
for(int t:nums){
FEntity f;
if(map.containsKey(t)){
f = map.get(t);
f.frequent++;
}else{
f = new FEntity(t, 1);
}
map.put(t, f);
}
//对hashmap的value 按照频率进行排序
List<FEntity> values = new ArrayList<FEntity>();
Set<Integer> keys = map.keySet();
for(Integer key:keys){
values.add(map.get(key));
}
Collections.sort(values);
//统计频率最高的k个数
for(int i=0;i<k;i++){
res.add(values.get(values.size()-1-i).num);
}
return res;
}

class FEntity implements Comparable<FEntity> {
int num, frequent;

public FEntity(int num, int frequent) {
this.num = num;
this.frequent = frequent;
}

public int compareTo(FEntity arg0) {
if (frequent > arg0.frequent)
return 1;
else if (frequent == arg0.frequent)
return 0;
else
return -1;
}
}