LeetCode-347. Top K Frequent Elementshttps://leetcode.com/problems/top-k-frequent-elements/

题目描述

Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.

Example 1:

Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]

Example 2:

Input: nums = [1], k = 1
Output: [1]

解题思路

【C++】

1. map + vector sort

class Solution {
public:vector<int> topKFrequent(vector<int>& nums, int k) {unordered_map<int,int> um;vector <int> res;for (int i=0;i<nums.size();i++) um[nums[i]]++;vector<pair<int,int>> freq_arr(um.begin(), um.end()); sort(freq_arr.begin(),freq_arr.end(),compare); for (int i = 0; i<k; i++) res.push_back(freq_arr[i].first);return res;}static bool compare(pair<int, int> p1, pair<int, int> p2) { if (p1.second == p2.second) return p1.first > p2.first; return p1.second > p2.second;     }
};

2. priority_queue + map

class Solution {
public:vector<int> topKFrequent(vector<int>& nums, int k) {auto cmp = [](const pair<int,int>& p1, const pair<int,int>& p2) {return p1.second > p2.second;};priority_queue<pair<int,int>, vector<pair<int,int>>, decltype(cmp)> q(cmp);unordered_map<int, int> m;for (int i : nums) m[i]++;for (auto& p : m) {if (q.size() < k) q.push(p);else {if (cmp(q.top(), p)) continue;q.push(p);q.pop();}}vector<int> results;for (; !q.empty(); results.push_back(q.top().first), q.pop());return results;}
};

3. 桶排序

class Solution {
public:vector<int> topKFrequent(vector<int>& nums, int k) {unordered_map<int, int> counts;int max_count = 0;for (const int & num : nums) {max_count = max(max_count, ++counts[num]);}vector<vector<int> > buckets(max_count + 1);for (const auto & p : counts) {buckets[p.second].push_back(p.first);}vector<int> res;for (int i = max_count; i >= 0 && res.size() < k; --i) {for (const int & num : buckets[i]) {res.push_back(num);if (res.size() == k) {break;}}}return res;}
};

【Java】

1. HashMap + PriorityQueue

class Solution {public int[] topKFrequent(int[] nums, int k) {Map<Integer,Integer> map= new HashMap<>();for (int a: nums) {map.put(a, map.getOrDefault(a, 0) + 1);}PriorityQueue<Map.Entry<Integer, Integer>> queue =new PriorityQueue<>((a,b) -> b.getValue() - a.getValue());for (Map.Entry<Integer, Integer> entry: map.entrySet()){queue.offer(entry);}int[] result = new int[k];for (int i=0; i<k; i++) {result[i] = queue.remove().getKey();}return result;}
}

2. 桶排序

class Solution {public int[] topKFrequent(int[] nums, int k) {HashMap<Integer, Integer> map = new HashMap<>();int maxCount = 0;for (int i=0; i<nums.length; i++) {map.put(nums[i], map.getOrDefault(nums[i], 0) + 1);maxCount = Math.max(maxCount, map.get(nums[i]));}List<Integer>[] bucket = new List[maxCount + 1];for (int e: map.keySet()) {int freq = map.get(e);if(bucket[freq] == null) {bucket[freq] = new ArrayList<>();}bucket[freq].add(e);}int[] ans = new int[k];int count = 0;for (int i=maxCount; i >= 0 && count < k; i--) {if (bucket[i] == null) {continue;}for (int num : bucket[i]) { ans[count++] = num;if(count == k) break;}}return ans;                   }
}

参考文献

【1】PriorityQueue的用法和底层实现原理_Hanniboo's Blog-CSDN博客_priorityqueue

【2】你应该知道的 PriorityQueue ——深入浅出分析 PriorityQueue_Java极客技术-CSDN博客

【3】C++中priority_queue理解与使用_不积跬步无以至千里-CSDN博客_c++ priority_queue

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