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.
題意:返回數組中出現次數前K高的元素:
public class Solution {
public List<Integer> topKFrequent(int[] array, int k) {
HashMap<Integer,Integer> hm = new HashMap<Integer,Integer>();
List<Integer> Ls = new ArrayList<Integer>();
int value=0;
for(int i = 0; i < array.length; i++){
if(hm.containsKey(array[i])){
hm.put(array[i],hm.get(array[i])+1 );
}else{
hm.put(array[i], 1);
}
}
List<Map.Entry<Integer, Integer>> entryList = new ArrayList<Map.Entry<Integer, Integer>>(hm.entrySet());
for (int i = 0; i < entryList.size(); i++) {
System.out.println(entryList.get(i));
}
Collections.sort(entryList, new Comparator<Map.Entry<Integer, Integer>>() {
public int compare(Map.Entry<Integer, Integer> o1, Map.Entry<Integer, Integer> o2) {
return (o2.getValue() - o1.getValue());//根據value排序
//return (o1.getKey()).toString().compareTo(o2.getKey());
}
});
for (int i = 0; i < k; i++) {
Ls.add(entryList.get(i).getKey());
}
return Ls;
}
}
第一獨立完成那麼長的代碼,雖然過程有點麻煩,也上網查了資料,但是這個是我自己寫出來的啊。效率不太高,OJ測試時耗時102ms,想想 就是一個長時間。