leetcode 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.

这一题我的思路就是用一个map统计每个元素出现的频率，再使用bucket sort对频率排序。
package leetcode;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
import java.util.Map;

public class Top_K_Frequent_Elements_347 {

public List<Integer> topKFrequent(int[] nums, int k) {
List<Integer> result=new ArrayList<Integer>();
HashMap<Integer, Integer> num_frequency=new HashMap<Integer, Integer>();
int maxFrequency=0;
for(int i=0;i<nums.length;i++){
int num=nums[i];
int frequency=num_frequency.getOrDefault(num, 0);
frequency++;
num_frequency.put(num,frequency);
maxFrequency=maxFrequency>frequency?maxFrequency:frequency;
}
String[] bucket=new String[maxFrequency+1];
Iterator iter = num_frequency.entrySet().iterator();
while (iter.hasNext()) {
Map.Entry entry = (Map.Entry) iter.next();
int num = (int)entry.getKey();
int frequency = (int)entry.getValue();
if(bucket[frequency]==null){
bucket[frequency]=num+"";
}
else{
bucket[frequency]=bucket[frequency]+","+num;
}
}
int theFrequency=maxFrequency;
while(k>0){
String theString=bucket[theFrequency];
if(theString!=null){
String[] splitWords=theString.split(",");
for(int i=0;i<splitWords.length&&k>0;i++){
result.add(Integer.parseInt(splitWords[i]));
k--;
}
}
theFrequency--;
}
return result;
}

public static void main(String[] args) {
// TODO Auto-generated method stub
Top_K_Frequent_Elements_347 t=new Top_K_Frequent_Elements_347();
int k=2;
int[] nums=new int[]{1,2};
List<Integer> list=t.topKFrequent(nums, k);
for(Integer i:list){
System.out.println(i);
}
}

}有大神的解法跟我一毛一样，只不过bucket数组的类型换了。讲道理我当初也想以list为数组的，但是不会写语法，这个大神给了我样例。
public List<Integer> topKFrequent(int[] nums, int k) {

List<Integer>[] bucket = new List[nums.length + 1];
Map<Integer, Integer> frequencyMap = new HashMap<Integer, Integer>();

for (int n : nums) {
frequencyMap.put(n, frequencyMap.getOrDefault(n, 0) + 1);
}

for (int key : frequencyMap.keySet()) {
int frequency = frequencyMap.get(key);
if (bucket[frequency] == null) {
bucket[frequency] = new ArrayList<>();
}
bucket[frequency].add(key);
}

List<Integer> res = new ArrayList<>();

for (int pos = bucket.length - 1; pos >= 0 && res.size() < k; pos--) {
if (bucket[pos] != null) {
res.addAll(bucket[pos]);
}
}
return res;
}除了桶排序，还可以使用最大堆PriorityQueue和红黑树TreeMap，其中TreeMap的性能是最好的。

优先级队列：
public class Solution {
public List<Integer> topKFrequent(int[] nums, int k) {
HashMap<Integer, Integer> mp = new HashMap<>();
for (int val: nums) {
mp.put(val, mp.getOrDefault(val, 0) + 1);
}
PriorityQueue<Map.Entry<Integer, Integer>> q = new PriorityQueue<>((a, b) -> (b.getValue() - a.getValue()));//最大堆
for (Map.Entry<Integer, Integer> entry : mp.entrySet()) {
q.add(entry);
}
List<Integer> ans = new ArrayList<>();
while (ans.size() < k) {
Map.Entry<Integer, Integer> entry = q.poll();
ans.add(entry.getKey());
}
return ans;
}
}
TreeMap:
public class Solution {
public List<Integer> topKFrequent(int[] nums, int k) {
HashMap<Integer, Integer> mp = new HashMap<>();
for (int val: nums) {
mp.put(val, mp.getOrDefault(val, 0) + 1);
}
TreeMap<Integer, List<Integer>> freqMap = new TreeMap<>();
for (int val: mp.keySet()) {
int cur = mp.get(val);
if (!freqMap.containsKey(cur)) {
freqMap.put(cur, new ArrayList<>());
}
freqMap.get(cur).add(val);
}
List<Integer> ans = new ArrayList<>();
while (ans.size() < k) {
//The pollLastEntry() method is used to remove and returns a key-value mapping
//associated with the greatest key in this map, or null if the map is empty.
Map.Entry<Integer, List<Integer>> entry = freqMap.pollLastEntry();
ans.addAll(entry.getValue());
}
return ans;
}
}