摘要:先按照元素次數(shù)的將所有元素存入,再按照次數(shù)元素將哈希表里的所有元素存入,然后取最后的個(gè)元素返回。
Problem
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.
Store each nums element and its count in HashMap.
Traverse its keySet(), store the count of each key into TreeMap, which means reverse the key-value pairs. And TreeMap will sort the elements by count value.
Use pollLastEntry() to get last K entries from TreeMap; then addAll() to put values in ArrayList res.
先按照元素-次數(shù)的pair將所有元素存入HashMap,再按照次數(shù)-元素pair將哈希表里的所有元素存入TreeMap,然后取TreeMap最后的k個(gè)元素返回。
public class Solution { public ListMin Heap lambda-expressiontopKFrequent(int[] nums, int k) { Map map = new HashMap<>(); for (int num: nums) { map.put(num, map.getOrDefault(num, 0)+1); } TreeMap > sorted = new TreeMap<>(); for (int num: map.keySet()) { int count = map.get(num); if (!sorted.containsKey(count)) sorted.put(count, new LinkedList<>()); sorted.get(count).add(num); } List res = new ArrayList<>(); while (res.size() < k) { Map.Entry > entry = sorted.pollLastEntry(); res.addAll(entry.getValue()); } return res; } }
public class Solution { public Listno-lambdatopKFrequent(int[] nums, int k) { List res = new ArrayList<>(); if (nums.length < k) return res; Map map = new HashMap<>(); for (int num: nums) { map.put(num, map.getOrDefault(num, 0)+1); } PriorityQueue > minHeap = new PriorityQueue<>((a, b)->(a.getValue()-b.getValue())); for (Map.Entry entry: map.entrySet()) { minHeap.offer(entry); if (minHeap.size() > k) minHeap.poll(); } while (res.size() < k) { res.add(minHeap.poll().getKey()); } return res; } }
public class Solution { public ListMax HeaptopKFrequent(int[] nums, int k) { List res = new ArrayList<>(); if (nums.length < k) return res; Map map = new HashMap<>(); for (int num: nums) { map.put(num, map.getOrDefault(num, 0)+1); } PriorityQueue > minHeap = new PriorityQueue >(new Comparator >() { public int compare(Map.Entry a, Map.Entry b) { return a.getValue()-b.getValue(); } }); for (Map.Entry entry: map.entrySet()) { minHeap.offer(entry); if (minHeap.size() > k) minHeap.poll(); } while (res.size() < k) { Map.Entry entry = minHeap.poll(); res.add(entry.getKey()); } return res; } }
public class Solution { public ListtopKFrequent(int[] nums, int k) { List res = new ArrayList<>(); if (nums.length < k) return res; Map map = new HashMap<>(); for (int num: nums) { map.put(num, map.getOrDefault(num, 0)+1); } PriorityQueue > maxHeap = new PriorityQueue<>((a, b) -> (b.getValue()-a.getValue())); for (Map.Entry entry: map.entrySet()) { maxHeap.offer(entry); } while (res.size() < k) { res.add(maxHeap.poll().getKey()); } return res; } }
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