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本帖最后由 Seawolf 于 2019-9-10 03:40 编辑
- Given a non-empty array of integers, return the k most frequent elements.
- Example 1:
- Input: nums = [1,1,1,2,2,3], k = 2
- Output: [1,2]
- Example 2:
- Input: nums = [1], k = 1
- Output: [1]
- 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.
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using regular hashmap!
- import java.util.Collection;
- class Solution {
- public List<Integer> topKFrequent(int[] nums, int k) {
- List<Integer> list = new ArrayList<>();
- HashMap <Integer,Integer> map = new HashMap<>();
- for(int i = 0; i< nums.length; i++){
- map.put(nums[i], map.getOrDefault(nums[i],0)+1);
- }
- Collection<Integer> c = map.values();
- Object[] obj = c.toArray();
- Arrays.sort(obj);
- int i = 0;
- while(k-- != 0){
- for(Integer key : map.keySet()){
- if (map.get(key).equals(Integer.parseInt(obj[obj.length- 1-i].toString())))
- {
-
- i++;
- list.add(key);
- map.remove(key);
- break;
- }
- }
- }
- return list;
- }
- }
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using bucket sort!!
- class Solution {
- public List<Integer> topKFrequent(int[] nums, int k) {
- Map<Integer,Integer> map = new HashMap<>();
- for(int a : nums) map.put(a,map.getOrDefault(a,0)+1);
- List<Integer> [] bucket = new List[nums.length + 1];
-
- for(Integer key : map.keySet()){
-
- int fre = map.get(key);
- if(bucket[fre] == null) bucket[fre] = new ArrayList<>();
- bucket[fre].add(key);
- }
- List<Integer> re = new ArrayList<>();
-
- for(int i = bucket.length - 1; i >=0 && re.size() < k ; i--){
-
- if(bucket[i] != null) re.addAll(bucket[i]);
- }
-
- return re;
-
- }
-
- }
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using java heap!
- import java.util.*;
- class Solution {
-
- class Pair{
- int num;
- int count;
- public Pair(int num, int count){
-
- this.num = num;
- this.count =count;
- }
- }
- public List<Integer> topKFrequent(int[] nums, int k) {
-
- Map<Integer, Integer> map = new HashMap<>();
-
- for(int a: nums) map.put(a,map.getOrDefault(a,0)+1);
-
- PriorityQueue<Pair> queue = new PriorityQueue<Pair>(new Comparator<Pair>(){
-
- public int compare(Pair a, Pair b){
-
- return a.count- b.count;
- }
- });
-
- for(Map.Entry<Integer,Integer> entry: map.entrySet()){
-
- Pair p = new Pair(entry.getKey(), entry.getValue());
-
- queue.offer(p);
- if(queue.size() > k) queue.poll();
- }
-
- List<Integer> list = new ArrayList<>();
-
- while(queue.size()>0) list.add(queue.poll().num);
-
- Collections.reverse(list);
-
- return list;
- }
-
- }
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