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 logn), where
n is the array's size.
这个题不难，就是复习下treemap的用法。

下面的工具类可以将Hashmap排序。将Hashmap转变成LinkedHashMap。

class MapUtil
{
public static <K, V extends Comparable<? super V>> Map<K, V>
sortByValue( Map<K, V> map )
{
List<Map.Entry<K, V>> list =
new LinkedList<Map.Entry<K, V>>( map.entrySet() );
Collections.sort( list, new Comparator<Map.Entry<K, V>>()
{
public int compare( Map.Entry<K, V> o1, Map.Entry<K, V> o2 )
{
return (o1.getValue()).compareTo( o2.getValue() );
}
} );

Map<K, V> result = new LinkedHashMap<K, V>();
for (Map.Entry<K, V> entry : list)
{
result.put( entry.getKey(), entry.getValue() );
}
return result;
}
}

AC代码：（当然可以用上面的方法做）

public class Solution {
class myComparator implements Comparator<Integer>{
Map<Integer, Integer> map;

public myComparator(Map<Integer, Integer> map) {
this.map=map;
}
@Override
public int compare(Integer o1, Integer o2) {
//大于，-1是降序，1是升序
if(map.get(o1)>map.get(o2))
return -1;
return 1;
}
}
public List<Integer> topKFrequent(int[] nums, int k) {
List<Integer> result=new ArrayList<Integer>();
Map<Integer, Integer> map=new TreeMap<Integer, Integer>();
for(int i=0;i<nums.length;i++){
if(!map.containsKey(nums[i])){
map.put(nums[i], 1);
}else{
map.put(nums[i], map.get(nums[i])+1);
}
}
Comparator<Integer> myComparator=new myComparator(map);
Map<Integer, Integer> treemap=new TreeMap<Integer, Integer>(myComparator);
treemap.putAll(map);
for (Integer key : treemap.keySet()) {
result.add(key);
k--;
if(k==0){
break;
}
}
return result;
}
}