azl397985856 commented 3 years ago

347. 前 K 个高频元素

入选理由

暂无

题目地址

https://leetcode-cn.com/problems/top-k-frequent-elements/

前置知识

哈希表
堆排序

快速选择

题目描述


给定一个非空的整数数组，返回其中出现频率前  k  高的元素。
示例 1:

输入: nums = [1,1,1,2,2,3], k = 2 输出: [1,2]

示例 2:

输入: nums = [1], k = 1 输出: [1]

提示：

你可以假设给定的 k 总是合理的，且 1 ≤ k ≤ 数组中不相同的元素的个数。
你的算法的时间复杂度必须优于 O(n log n) , n 是数组的大小。
题目数据保证答案唯一，换句话说，数组中前 k 个高频元素的集合是唯一的。
你可以按任意顺序返回答案。

来源：力扣（LeetCode）链接：https://leetcode-cn.com/problems/top-k-frequent-elements 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

HackBL commented 3 years ago

Map + heap

class Solution {
public int[] topKFrequent(int[] nums, int k) {
    int[] res = new int[k];
    Map<Integer, Integer> map = new HashMap<>();
    for (int n: nums) {
        map.put(n, map.getOrDefault(n, 0)+1);
    }

    PriorityQueue<Map.Entry<Integer, Integer>> heap = new PriorityQueue<>((a,b) -> a.getValue() - b.getValue());

    for (Map.Entry<Integer,Integer> entry: map.entrySet()) {
        heap.offer(entry);

        if (heap.size() > k) {
            heap.poll();
        }
    }

    for (int i = 0; i < k; i++) {
        Map.Entry<Integer, Integer> entry = heap.poll();
        res[i] = entry.getKey();
    }

    return res;
}
}

Time: O(nlogk)
Space: O(n + k)

Tomtao626 commented 3 years ago

思路

小顶堆实现队列

代码


import heapq
class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
#要统计元素出现频率
map_ = {} #nums[i]:对应出现的次数
for i in range(len(nums)):
map_[nums[i]] = map_.get(nums[i], 0) + 1

#对频率排序
#定义一个小顶堆，大小为k
pri_que = [] #小顶堆

#用固定大小为k的小顶堆，扫面所有频率的数值
for key, freq in map_.items():
    heapq.heappush(pri_que, (freq, key))
    if len(pri_que) > k: #如果堆的大小大于了K，则队列弹出，保证堆的大小一直为k
        heapq.heappop(pri_que)

#找出前K个高频元素，因为小顶堆先弹出的是最小的，所以倒叙来输出到数组
result = [0] * k
for i in range(k-1, -1, -1):
    result[i] = heapq.heappop(pri_que)[1]
return result


## 复杂度

时间：O(nlogk)
空间：O(n)

yanglr commented 3 years ago

思路

堆排序 + 哈希表

topK问题, 容易想到使用优先队列(堆排序)。

代码

实现语言: C++

class Solution {
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        vector<int> res;
        unordered_map<int, int> dict; // 映射: value -> count
        for (auto num : nums)
        {
            if (dict.find(num) == dict.end()) dict[num] = 0;
            else dict[num]++;
        }
        auto cmp = [](pair<int, int> p1, pair<int, int> p2)
        {
            return p1.second < p2.second;  /* 大顶堆 */
        };

        priority_queue<pair<int, int>, vector<pair<int, int>>, decltype(cmp)> q(cmp);
        for (auto kvp : dict) q.push(kvp);
        while (k > 0)
        {
            res.push_back(q.top().first);
            q.pop();

            k--;
        }

        return res;
    }
};

复杂度分析

时间复杂度：O(k logm), m 是数组中不同数字的数量，堆中有 m 个元素
空间复杂度：O(m), m 是数组中不同数字的数量

st2yang commented 3 years ago

思路

Counter + bucket

代码

python

class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
    n = len(nums)
    counter = collections.Counter(nums)
    buckets = [[] for _ in range(n)]
    for key in counter:
        buckets[counter[key]-1].append(key)

    res=[]
    for i in range(n - 1, -1, -1):
        for b in buckets[i]:
            res.append(b)
            if len(res) >= k:
                return res
    return res

复杂度

时间: O(n)
空间: O(n)

BpointA commented 3 years ago

思路

暴力解哈希表+排序，时间复杂度为nlogn，采用堆之后达到进阶要求。

Python3代码

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        from collections import defaultdict
        import heapq
        d=defaultdict(int)
        for i in nums:
            d[i]+=1
        kv=list(d.items())
        res=[]
        for i in range(k):
            heapq.heappush(res,[kv[i][1],kv[i][0]])

        for j in range(k,len(kv)):
            k,v=kv[j]
            if v>res[0][0]:
                heapq.heappop(res)
                heapq.heappush(res,[kv[j][1],kv[j][0]])

        final=[i[1] for i in res]
        return final

复杂度

时间复杂度：O(nlogk)，堆排序

空间复杂度：O(n)，哈希表长度

附录

暴力解法。

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        from collections import defaultdict
        d=defaultdict(int)
        for i in nums:
            d[i]+=1
        r=sorted(d.items(),key=lambda x:-x[1])
        res=[]
        for i in range(k):
            res.append(r[i][0])
        return res

BreezePython commented 3 years ago

思路

哈希表+排序

代码

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        dic = Counter(nums)
        return sorted(dic.keys(),key = lambda x:-dic[x])[:k]

复杂度：

时间复杂度：O(nlog(n))
空间复杂度：O(m) m为数组中不同的元素数目

CoreJa commented 3 years ago

思路

只想到了用哈希表然后排序的方法，之后学习一下堆排的写法

代码

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        d = {}
        for num in nums:
            if num not in d:
                d[num] = 1
            else:
                d[num] += 1
        t: List[Tuple] = sorted(d.items(), key=lambda x: x[1], reverse=True)
        return [i[0] for i in t[:k]]

Bochengwan commented 3 years ago

思路

使用hashmap存frequency，O(N).

然后利用heap得到topk的element，O(KLOGN)

代码

import collections, heapq as hp
class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        my_list = collections.defaultdict(int)
        for num in nums:
            my_list[num] += 1
        result = []
        b = []
        for key in my_list:
            b.append((-my_list[key],key))
        hp.heapify(b)
        result = []
        for i in range(k):
            result.append(hp.heappop(b)[1])
        return result

复杂度分析

时间复杂度：O(klogn)，其中 N 为数组长度。
空间复杂度：O(n)

thinkfurther commented 3 years ago

思路

用dict记录每一个数的频率，然后排序

代码

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        import collections
        count_dict = collections.defaultdict(int)
        for i in nums:
            count_dict[i] += 1

        sorted_dict = sorted(count_dict.items(), key = lambda x:x[1],reverse = True)
        result = []

        for i in range(k):
            result.append(sorted_dict[i][0])
        return result

复杂度

时间复杂度：O(N*logN)

空间复杂度：O(N)

harleyszhang commented 3 years ago

解题思路

哈希表 unordered_map + 最小堆 priority_queue 的方法（C++）

class Solution {
public:
   // 小顶堆
    class mycomparison {
    public:
        bool operator()(const pair<int, int>& lhs, const pair<int, int>& rhs) {
            return lhs.second > rhs.second;
        }
    };

    vector<int> topKFrequent(vector<int>& nums, int k) {
        unordered_map<int, int> map;
        // 创建 map，map的键是数组元素，键值是数组元素出现次数
        for(auto num: nums){
            map[num]++;
        }

        // 定义一个小顶堆，并维持大小为 k
        priority_queue<pair<int, int>, vector<pair<int, int>>, mycomparison > heap; // 小顶堆，堆顶为最小值
        for (unordered_map<int, int>::iterator it = map.begin(); it != map.end(); it++) {
            heap.push(*it);
            if(heap.size() > k) heap.pop(); // 弹出堆顶元素，即最小元素
        }
        // heap 元素倒序输出到数组
        vector<int> ret(k);
        for (int i=k-1; i>=0; i--){
            ret[i] = heap.top().first;
            heap.pop();
        }
        return ret;
    }
};

Mahalasu commented 3 years ago

思路

priority queue + hashmap

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        if k == len(nums):
            return nums

        count = {}
        for item in nums:
            if item not in count:
                count[item] = (0, item)
            count[item] = (count[item][0] + 1, item)

        heap = []
        for value in count.values():
            heapq.heappush(heap, value)
            if len(heap) > k:
                heapq.heappop(heap)

        ans = []
        for i in range(k):
            ans.append(heapq.heappop(heap)[1])

        return ans

T: O(nlogk) S: O(n + k)

yachtcoder commented 3 years ago

The power of Counter. Time: O(nlgk) because most_common use a heap of size k to find the most frequent k items. Space: O(n)

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        return [x[0] for x in Counter(nums).most_common(k)]

kidexp commented 3 years ago

thoughts

先拿到nums里面每个num的count

然后把他放到最大queue里面

然后pop k次放进results

code

import heapq
from collections import Counter

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        """
        heap solution
        """
        num_count_dict = Counter(nums)
        num_count_heap = [(-count, num) for num, count in num_count_dict.items()]
        heapq.heapify(num_count_heap)
        result = []
        for _ in range(k):
            result.append(heapq.heappop(num_count_heap)[1])
        return result

complexity

time O(klgn)

space O(n)

SunnyYuJF commented 3 years ago

思路

Heap + Counter

代码 Python

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        Counter = collections.Counter(nums)  #O(N)
        if len(Counter)<=k:
            return Counter.keys()
        heap=[]
        for key,value in Counter.items(): #O(NlogK)
            heapq.heappush(heap,(value,key))
            if len(heap)>k:
                heapq.heappop(heap)

        return [key for value,key in heap]

复杂度分析

时间复杂度：O(NlogK)
空间复杂度: O(N+K)

florenzliu commented 3 years ago

Explanation

Approach 1: use a hashmap to store all the numbers and count its frequency, then sort by its frequency and return the first k keys
Approach 2: use a heap, return the first k elements

Python

# Approach 1
class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        result = []
        d = defaultdict(int)
        for n in nums:
            d[n] += 1
        for i in sorted(d.items(), key=lambda item: item[1], reverse=True):
            if k == 0:
                return result
            result.append(i[0])
            k -= 1
        return result

# Approach 2

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        if k == len(nums):
            return nums
        count = Counter(nums)
        return heapq.nlargest(k, count.keys(), key=count.get)

Complexity

Approach 1:

Time Complexity: O(N)
Space Complexity: O(N)

Approach 2:

Time complexity: O(Nlogk) if k < N and O(N) if k = N
Space complexity: O(N + k)

tongxw commented 3 years ago

思路

哈希表计数，根据计数大小，对哈希表的keySet（所有不同数字）排序后输出前k个元素

代码

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number[]}
 */
var topKFrequent = function(nums, k) {
    const counters = new Map();
    nums.forEach(num => {
      if (counters.has(num)) {
        counters.set(num, counters.get(num) + 1);
      } else {
        counters.set(num, 1);
      }
    });

    return [...counters.keys()].sort((n1, n2) => {
      return counters.get(n2) - counters.get(n1);
    }).slice(0, k);
};

TC: O(nlogn)
SC: O(nlogn) n为不同数字个数

nonevsnull commented 3 years ago

思路

拿到题的直觉解法
- topK，比较容易联想到heap
- 首先sort一下array
- 然后通过滚动的方式计算count，放入heap
- 最后poll出k个存入结果

AC

代码

class Solution {
    public int[] topKFrequent(int[] nums, int k) {
        Arrays.sort(nums);

        int count = 1;

        PriorityQueue<int[]> heap = new PriorityQueue<>((o1, o2)->o2[1] - o1[1]);
        for(int i = 0;i < nums.length;i++){
            if(i < nums.length - 1 && nums[i] == nums[i+1]){
                count++;
            } else {
                heap.add(new int[]{nums[i], count});
                count = 1;
            }
        }
        int[] res = new int[k];

        for(int i = 0;i < k;i++){
            res[i] = heap.poll()[0];
        }
        return res;
    }
}

复杂度

time: 排序,O(NlogN)，遍历一次求count，O(N)，弹出topK,O(KlogD), D是不同的元素的个数，最坏为O(KlogN);总的复杂度取最高O(NlogN); space：heap所需的空间O(D), D是不同的元素的个数，最坏为O(N)

cicihou commented 3 years ago


class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        dic = {}
        for i in nums:
            dic[i] = dic.get(i, 0) + 1

        tmp = {}
        for key, v in dic.items():
            tmp.setdefault(v, []).append(key)

        res = []
        for i in range(k):
            res += tmp.pop(max(tmp))
            if len(res) == k:
                return res

JiangyanLiNEU commented 3 years ago

Idea

use a dictionary to record the frequency of each num in nums.

use maxheap to maintain the order of frequency

Implement (Runtime=O(nlogn), Space=O(n) )

class Solution(object):
def topKFrequent(self, nums, k):
    dic = Counter(nums)
    maxheap = []
    heapq.heapify(maxheap)
    for key in dic.keys():
        heapq.heappush(maxheap, [-dic[key], key])

    result = []
    while k > 0 and maxheap:
        getOne = heapq.heappop(maxheap)
        result.append(getOne[1])
        k -= 1
    return result

heyqz commented 3 years ago

代码

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        buckets = [[] for _ in range(len(nums) + 1)]
        d = {}
        for num in nums:
            d[num] = d.get(num, 0) + 1

        for num, freq in d.items():
            buckets[freq].append(num)

        res = []
        for i in range(len(buckets)-1, -1, -1): 
            bucket = buckets[i]
            if bucket:
                for num in bucket:
                    res.append(num)
        return res[:k]

xj-yan commented 3 years ago

Quick Select

class Solution {
    public int[] topKFrequent(int[] nums, int k) {
        Map<Integer, Integer> counterMap = new HashMap<>();
        for (int n : nums){
            counterMap.put(n, counterMap.getOrDefault(n, 0) + 1);
        }

        int[] array = new int[counterMap.size()];
        int index = 0;
        for (int n : counterMap.keySet()){
            array[index++] = n;
        }

        quickSelect(array, 0, array.length - 1, counterMap, k);
        return Arrays.copyOfRange(array, array.length - k, array.length);
    }

    private void quickSelect(int[] array, int left, int right, Map<Integer, Integer> map, int k){
        int pivot = map.get(array[right]), index = left;

        for (int i = left; i < right; i++){
            if (map.get(array[i]) <= pivot){
                swap(array, i, index++);
            }
        }
        swap(array, index, right);
        int count = right - index + 1;
        if (count == k) return;
        else if (count > k) quickSelect(array, index + 1, right, map, k);
        else quickSelect(array, left, index - 1, map, k - count);
    }

    private void swap(int[] nums, int i, int j){
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
    }
}

Time Complexity: Best Case: O(N), Worst Case: O(N^2) Space Complexity: O(N)

pophy commented 3 years ago

思路

遍历数组用hashmap记录每个数出现的次数
利用大小为K的小根堆排序hashmap中的entry

Java code

    public int[] topKFrequent(int[] nums, int k) {
        Map<Integer, Integer> countMap = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            countMap.put(nums[i], countMap.getOrDefault(nums[i], 0) + 1);
        }
        PriorityQueue<Map.Entry<Integer, Integer>> priorityQueue = new PriorityQueue<>(Comparator.comparingInt(Map.Entry::getValue));
        for (Map.Entry<Integer, Integer> entry : countMap.entrySet()) {
            priorityQueue.add(entry);
            if (priorityQueue.size() > k) {
                priorityQueue.poll();
            }
        }
        List<Integer> resList = new ArrayList<>();
        while (!priorityQueue.isEmpty()) {
            resList.add(priorityQueue.poll().getKey());
        }
        int[] res = new int[resList.size()];
        for (int i = 0; i < res.length; i++) {
            res[i] = resList.get(i);
        }
        return res;
    }

时间&空间

时间 o(n*log(n)) 小根堆插入
空间 O(n)

falconruo commented 3 years ago

思路:

使用hashmap存放数组的值和frequency
由大到小按照frequency排序:
- 方法一:传统sort排序
- 方法二:小顶堆(Min Heap),存放最多k个pair

复杂度分析: 方法一、时间复杂度: O(nlogn), n为数组nums的元素个数空间复杂度: O(n + k)

方法二时间复杂度: O(nlogk), n为数组nums的元素个数空间复杂度: O(n + k)

代码(C++):


//方法二:
class Solution {
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        map<int, int> freq;

        // use a hashmap to store the frequency of each element
        for (auto n : nums) // O(n)
           freq[n]++;

        // use min heap to store the k most frequent elements

        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;

        for (auto f : freq) { // o(nlogk)
            if (pq.size() < k)
                pq.push({f.second, f.first}); // stack sort o(logk)
            else if (pq.top().first < f.second) {
                pq.pop();
                pq.push({f.second, f.first});
            }
        }

        vector<int> res;

        while (!pq.empty()) { // O(k)
            res.push_back(pq.top().second);
            pq.pop();
        }

        return res;
    }
};

MonkofEast commented 3 years ago

347. Top K Frequent Elements

Click Me

Algo

hashmap, key->number, val->frequency

Heap sort

heap to arr

Code

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        # corner case
        if k == len(nums): return nums
        # collections.Counter is a hashmap that key->number, val->frequence
        count = collections.Counter(nums)
        # heapq.nlargest(k, count.keys(), key=count.get)
        # means "find k largest" + "from count.keys()" + "by count.get(akey)"
        return heapq.nlargest(k, count.keys(), key=count.get)

Comp

k is the number of n-largest

T: O(Nlogk)

S: O(N+k)

chakochako commented 3 years ago

def topKFrequent(self, nums: List[int], k: int) -> List[int]:
    count = collections.Counter(nums)
    return [item[0] for item in count.most_common(k)]

yingliucreates commented 3 years ago

link:

https://leetcode.com/problems/top-k-frequent-elements/

代码 Javascript

const topKFrequent = (nums, k) => {
  const map = {};
  const result = [];
  const bucket = Array(nums.length + 1)
    .fill()
    .map(() => []);

  for (let num of nums) {
    map[num] = ~~map[num] + 1;
  }

  for (let num in map) {
    bucket[map[num]].push(parseInt(num));
  }

  for (let i = nums.length; i >= 0 && k > 0; k--) {
    while (bucket[i].length === 0) i--;
    result.push(bucket[i].shift());
  }

  return result;
};

复杂度分析

空间时间均为O(n)

pan-qin commented 3 years ago

idea: use a hashmap to record the frequency of each number. Then construct a min-Heap with size of k. The min-heap hold the entryset of the map. When it's size is larger than k, poll the top of the heap. In the end, output the whole heap. Time: O(NlogK) Space: O(n)

class Solution {
    public int[] topKFrequent(int[] nums, int k) {
        HashMap<Integer, Integer> counter = new HashMap<>();
        for(int num: nums)
            counter.put(num,counter.getOrDefault(num,0)+1);
        PriorityQueue<Map.Entry<Integer,Integer>> pq = new PriorityQueue<>((a,b)->a.getValue()-b.getValue());
        for(Map.Entry<Integer, Integer> entry: counter.entrySet()) {
            pq.offer(entry);
            if(pq.size()>k)
                pq.poll();
        }
        int[] res = new int[k];
        int i=0;
        while(!pq.isEmpty()) {
            res[i++]=pq.poll().getKey();
        } 
        return res;
    }
}

RocJeMaintiendrai commented 3 years ago

题目

https://leetcode-cn.com/problems/top-k-frequent-elements/

思路

HashMap记录数字的频率，使用一个size为K的最小堆来记录，堆顶元素为前k个高频中频率最低的那个，当新的数字来时只需要对比两个的频率即可。

代码

class Solution {
    public int[] topKFrequent(int[] nums, int k) {
        Map<Integer, Integer> map = new HashMap<>();
        for(int num : nums) {
            if(map.containsKey(num)) {
                map.put(num, map.get(num) + 1);
            } else {
                map.put(num, 1);
            }
        }

        Queue<Integer> queue = new PriorityQueue<>((a, b) -> map.get(a) - map.get(b));
        for(Integer num : map.keySet()) {
            if(queue.size() < k) {
                queue.offer(num);
            } else if(map.get(num) > map.get(queue.peek())) {
                queue.poll();
                queue.offer(num);
            }
        }

        int[] res = new int[k];
        for(int i = 0; i < k; i++) {
            res[i] = queue.poll();
        }
        return res;
    }
}

复杂度分析

时间复杂度

O(n) + O (nlogk)

空间复杂度

O(n + k)

mmboxmm commented 3 years ago

思路

重点题，bucket 和 quickSelect都要练习一下

代码

fun topKFrequent(nums: IntArray, k: Int): IntArray {
  val map = nums.asSequence().groupingBy { it }.eachCount()
  val bucket = Array<MutableList<Int>?>(nums.size + 1) { null }

  map.forEach {
    bucket[it.value] = bucket[it.value]?.apply {
      add(it.key)
    } ?: mutableListOf(it.key)
  }

  val res = mutableListOf<Int>()
  for (freq in bucket.lastIndex downTo 1) {
    bucket[freq]?.apply {
      res.addAll(take(minOf(k - res.size, size)))
    }
    if (res.size == k) return res.toIntArray()
  }
  return res.toIntArray()
}

复杂度

Time: O(n)

zol013 commented 3 years ago

思路：用hashmap记录数字的频率，然后将hashmap里的元素排序，然后返回前k个。 Python 3 Code:

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        hashmap = defaultdict(int)
        for num in nums:
            hashmap[num] += 1
        ans = []
        freqs = sorted(hashmap.items(), key = lambda x:x[1])
        while k > 0:
            num, freq = freqs.pop()
            ans.append(num)
            k -= 1
        return ans

Time Complexity: O(N*log(N) Space Complexity: O(N)

qixuan-code commented 3 years ago

LC 347. Top K Frequent Elements

Ideas

use hashmap to store frequency count
sort by frequency and get top k number

comment Runtime: 40 ms beat 75%

python代码

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:

        hashmap = dict()

        for num in nums:
            hashmap[num] = 0

        for num in nums:
            if num in hashmap:
                hashmap[num] +=1

        sort_map = dict(sorted(hashmap.items(), key = lambda item:item[1],reverse=True))
        i = 0
        result = []
        for num, count in enumerate(sort_map):
            if i < k:
                result.append(count)
                i = i+1
            else:
                break

        return result

复杂度分析

时间复杂度：O(nlogn + 2n + k)
空间复杂度：O(n)

bingyingchu commented 3 years ago


import heapq

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        freq = {}

        for num in nums:
            freq[num] = freq.get(num,0) + 1

        heap = []
        for num, count in freq.items():
            heapq.heappush(heap,(count,num))
            if len(heap) > k:
                heapq.heappop(heap)

        return [pairs[1] for pairs in heap ]

# Time: O(nlogk) / Space: O(n + k)

wangzehan123 commented 3 years ago


class Solution {
    public int[] topKFrequent(int[] nums, int k) {
        Map<Integer,Integer> map =new HashMap<>();
        int len = nums.length;
        int maxCount=1;
        for(int i=0;i<len;i++){
            int count = map.getOrDefault(nums[i],0)+1;
            if(count>maxCount){
                maxCount = count;
            }
            map.put(nums[i],count);
        }
        List<List<Integer>> list = new ArrayList<>(maxCount+1);
        for(int i=0;i<maxCount+1;i++){
            List<Integer> l = new ArrayList<>();
            list.add(l);
        }

        //遍历map,放到list中
        for(Integer key:map.keySet()){
           Integer count = map.get(key);
           List<Integer> mid = list.get(count);
           mid.add(key);
        }
        //从数组的后面开始遍历取k个
        int[] res = new int[k];
        int index=0;
        for(int i=maxCount;i>=0;i--){
            if(index>=k){
                break;
            }
            List<Integer> items = list.get(i);
            if(!items.isEmpty()){//说明有值
                for(Integer item:items){
                    res[index++]=item;
                }
            }
        }
        return res;
    }
}

复杂度分析

令 n 为数组长度。

时间复杂度：$O(n)$
空间复杂度：$O(n)$

ghost commented 3 years ago

题目

Top K Frequent Elements

思路

Hash table and bucket sort

代码

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        memo = collections.defaultdict(lambda:0)
        bucket = [[] for _ in range(len(nums)+1)] 
        res = []

        for num in nums:
            memo[num] += 1        

        for key in memo:
            bucket[memo[key]].append(key)

        for i in range(len(bucket)-1, 0, -1):

            if k <=0: break
            if bucket[i]:
                res+=bucket[i]
                k -= len(bucket[i])

        return res

复杂度

Space: O(N) Time: O(N)

Daniel-Zheng commented 3 years ago

思路

HashMap。

代码(C++)

class Solution {
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        unordered_map<int, int> hashMap;
        vector<pair<int, int>> count;
        vector<int> res;
        for (int i = 0; i < nums.size(); i++) {
            if (hashMap.count(nums[i])) {
                hashMap[nums[i]]++;
            } else {
                hashMap[nums[i]] = 1;
            }
        }
        for (auto &i : hashMap) count.push_back({i.second, i.first});
        sort(count.rbegin(), count.rend());
        for (int i = 0; i < k; i++) res.push_back(count[i].second);
        return res;
    }
};

复杂度分析

时间复杂度：O(N)
空间复杂度：O(N)

laofuWF commented 3 years ago

# use hashmap: key(num), value(-count)
# use priority queue to get top K frequent number
# time: O(NlogN), N is the number of unique elements in nums
# space: O(N)

import heapq

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        if not nums or not k:
            return []

        counter = {}
        for n in nums:
            counter[n] = counter.get(n, 0) - 1

        heap = []
        # push a pair into heap, sort by first element by default
        for key in counter:
            heapq.heappush(heap, (counter[key], key))

        res = []
        while k > 0 and heap:
            res.append(heapq.heappop(heap)[1])
            k -= 1

        return res

Menglin-l commented 3 years ago

思路：

小顶堆

代码部分：

class Solution {
    public int[] topKFrequent(int[] nums, int k) {
        // 先用map记住每一个元素出现的次数
        Map<Integer, Integer> count = new HashMap<>();
        for (int num : nums) {
            count.put(num, count.getOrDefault(num, 0) + 1);
        }

        // 自定义比较器，按出现次数升序排列
        PriorityQueue<Integer> pq = new PriorityQueue<>(k + 1, (a, b)->count.get(a) - count.get(b));

        // 填充队列
        for(int num : count.keySet()) {
            pq.add(num);
            if (pq.size() > k) pq.remove();
        }

        // 转换为数组并返回
        int[] ans = new int[k];
        int index = 0;
        while (!pq.isEmpty()) {
            ans[index ++] = pq.remove();
        }

        return ans;
    }
}

复杂度：

Time: O(Nlogk)

Space: O(N)

weichuliao commented 3 years ago

Main Idea

Code with Python

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        if k == len(nums):
            return nums

        count = collections.Counter(nums)

        return heapq.nlargest(k, count.keys(), key=count.get)

Complexity Analysis

Time:
Space:

SunStrongChina commented 3 years ago

347. 前 K 个高频元素

入选理由

暂无

题目地址

https://leetcode-cn.com/problems/top-k-frequent-elements/

前置知识

哈希表
堆排序
快速选择

题目描述

给定一个非空的整数数组，返回其中出现频率前  k  高的元素。
示例 1:

输入: nums = [1,1,1,2,2,3], k = 2
输出: [1,2]

示例 2:

输入: nums = [1], k = 1
输出: [1]

提示：

- 你可以假设给定的  k  总是合理的，且 1 ≤ k ≤ 数组中不相同的元素的个数。
- 你的算法的时间复杂度必须优于 O(n log n) , n  是数组的大小。
- 题目数据保证答案唯一，换句话说，数组中前 k 个高频元素的集合是唯一的。
- 你可以按任意顺序返回答案。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/top-k-frequent-elements
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路：使用字典记录元素的出现频率，然后按照频率排序，取前几个就行了

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        dict1={}
        for num1 in nums:
            if num1 not in dict1:
                dict1[num1]=1
            else:
                dict1[num1]+=1
        dict1_items=sorted(dict1.items(),key=lambda x:x[1],reverse=True)
        allRes=[]
        for i in range(k):
            allRes.append(dict1_items[i][0])
        return allRes

时间复杂度：O(N) 空间复杂度：O(N)

pzl233 commented 3 years ago

思路

用HashMap存频率，用PriorityQueue取出频率最高的k个元素

代码

class Solution {
    public int[] topKFrequent(int[] nums, int k) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int i : nums) {
            map.put(i, map.containsKey(i) ? map.get(i) + 1 : 1);
        }
        int[] result = new int[k];
        PriorityQueue<Integer> pq = new PriorityQueue<>( (a, b) -> map.get(b) - map.get(a));
        for (int i : map.keySet()) {
            pq.add(i);
        }
        for (int i = 0; i < k; i++) {
            result[i] = pq.poll();
        }
        return result;
    }
}

复杂度分析

时间复杂度： O(nlogn) 空间复杂度: O(n)

shawncvv commented 3 years ago

思路

官方解题思路

代码

JavaScript Code

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number[]}
 */
var topKFrequent = function (nums, k) {
  const counts = {};
  for (let num of nums) {
    counts[num] = (counts[num] || 0) + 1;
  }
  let pairs = Object.keys(counts).map((key) => [counts[key], key]);

  select(0, pairs.length - 1, k);
  return pairs.slice(0, k).map((item) => item[1]);

  // 快速选择
  function select(left, right, offset) {
    if (left >= right) {
      return;
    }
    const pivotIndex = partition(left, right);
    console.log({ pairs, pivotIndex });
    if (pivotIndex === offset) {
      return;
    }

    if (pivotIndex <= offset) {
      select(pivotIndex + 1, right, offset);
    } else {
      select(left, pivotIndex - 1);
    }
  }

  // 拆分数组为两个part
  function partition(left, right) {
    const [pivot] = pairs[right];
    let cur = left;
    let leftPartIndex = left;
    while (cur < right) {
      if (pairs[cur][0] > pivot) {
        swap(leftPartIndex++, cur);
      }
      cur++;
    }
    swap(right, leftPartIndex);
    return leftPartIndex;
  }

  function swap(x, y) {
    const term = pairs[x];
    pairs[x] = pairs[y];
    pairs[y] = term;
  }
};

复杂度

时间复杂度： O(N), 最坏能到 O(N^2)
空间复杂度: O(N)

xiaowenhe commented 3 years ago

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        counter = collections.Counter(nums)
        sorted_nums = sorted(counter.items(), key = lambda x: -x[1])
        result = [key for key, val in sorted_nums]
        return result[:k]

Francis-xsc commented 3 years ago

思路

哈希表+自定义排序

代码


class Solution {
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        unordered_map<int,int>m;
        for(auto x:nums)
            m[x]++;
        vector<pair<int,int>>v;
        for(pair<int,int> x:m)
            v.push_back(x);
        sort(v.begin(),v.end(),cmp);
        vector<int>res;
        for(int i=0;i<k;i++)
            res.push_back(v[i].first);
        return res;
    }
    static bool cmp(pair<int,int>a,pair<int,int>b)
    {
        return a.second>b.second;
    }
};

复杂度分析

时间复杂度：O(N)
空间复杂度：O(1)

liuyangqiQAQ commented 3 years ago

直接map存储排序

class Solution {
    public int[] topKFrequent(int[] nums, int k) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int num : nums) {
            map.put(num, map.getOrDefault(num, 0) + 1);
        }
        return new ArrayList<>(map.keySet()).stream().sorted((o1, o2) -> map.get(o2) - map.get(o1)).limit(k).mapToInt(Integer::intValue).toArray();
    }
}

flame0409 commented 3 years ago

class Solution {
    public List<Integer> topKFrequent(int[] nums, int k) {
        // 使用字典，统计每个元素出现的次数，元素为键，元素出现的次数为值
        HashMap<Integer,Integer> map = new HashMap();
        for(int num : nums){
            if (map.containsKey(num)) {
               map.put(num, map.get(num) + 1);
             } else {
                map.put(num, 1);
             }
        }
        // 遍历map，用最小堆保存频率最大的k个元素
        PriorityQueue<Integer> pq = new PriorityQueue<>(new Comparator<Integer>() {
            @Override
            public int compare(Integer a, Integer b) {
                return map.get(a) - map.get(b);
            }
        });
        for (Integer key : map.keySet()) {
            if (pq.size() < k) {
                pq.add(key);
            } else if (map.get(key) > map.get(pq.peek())) {
                pq.remove();
                pq.add(key);
            }
        }
        // 取出最小堆中的元素
        List<Integer> res = new ArrayList<>();
        while (!pq.isEmpty()) {
            res.add(pq.remove());
        }
        return res;
    }
}

时间复杂度：O(nlogn)

空间复杂度：O(n)

comst007 commented 3 years ago

347. 前K个高频元素

思路

先统计出每个元素的出现次数。
借助桶排序的思想从后往前遍历找出出现次数大于K的元素的最小出现频率。那原数组中
出现频率大于该最小频率的元素就是出现频率最高的K个元素。

代码

C++

class Solution {
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        unordered_map<int, int> h_cnt;
        for(auto& cur:nums){
            h_cnt[cur]++;
        }
        vector<int> cnt_arr(nums.size() + 1, 0);
        for(auto [cur, cnt]: h_cnt){
            cnt_arr[cnt] ++;
        }

        int ans_res = 0;
        int target_cnt, ii;
        for(target_cnt = nums.size(); target_cnt >=0; -- target_cnt){
            ans_res += cnt_arr[target_cnt];
            if(ans_res >= k){
                break;
            }
        }
        vector<int> ans;
        for(auto [cur, cnt]: h_cnt){
            if(cnt >= target_cnt){
                ans.push_back(cur);
                if(ans.size() == k) break;
            }
        }
        return ans;
    }
};

复杂度分析

n 为数组元素个数。

时间复杂度 O(n)
空间复杂度 O(n)

jaysonss commented 3 years ago

思路

建立哈希表，保存下每个数字出现的次数

对哈希值列表进行排序

关键点˜

无

代码

语言支持：Java

Java Code:

class Solution {
    public int[] topKFrequent(int[] nums, int k) {
        Map<Integer,int[]> map = new HashMap<>();
        for(int num : nums){
            int[] count = map.get(num);
            if(count == null){
                map.put(num,new int[]{num,1});
            }else{
                count[1]+=1;
            }
        }

        List<int[]> countList = new ArrayList<>(map.values());
        Collections.sort(countList,new Comparator<int[]>(){
            @Override
            public int compare(int[] o1, int[] o2) {
                return o2[1]-o1[1];
            }
        });
        int[] ans = new int[k];

        for(int i=0;i<k;i++){
            ans[i]=countList.get(i)[0];
        }
        return ans;
    }
}

复杂度分析

时间复杂度：O(nlogn)
空间复杂度：O(n)

septasset commented 3 years ago

思考

关键点

代码(Python)

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        c = Counter(nums)
        # freqs = c.items()
        # sorted(freqs, key=lambda x:-x[1])
        return [x[0] for x in c.most_common(k)]

复杂度分析

时间：O（nlogn）
空间：O（n）

hwpanda commented 3 years ago

const topKFrequent = (nums, k) => {
  let map = new Map()
  for (let i = 0; i < nums.length; i++) {
      if (map.has(nums[i])) {
          map.set(nums[i], map.get(nums[i]) + 1)
      }
      else {
          map.set(nums[i], 1)
      }
  }
  let res = [...map]
  res.sort((a, b) => b[1] - a[1])
  return res.map((n) => n[0]).slice(0, k)
};

Serena9 commented 3 years ago

代码

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        def sift_down(arr, root, k):
            val = arr[root] 
            while root<<1 < k:
                child = root << 1
                if child|1 < k and arr[child|1][1] < arr[child][1]:
                    child |= 1
                if arr[child][1] < val[1]:
                    arr[root] = arr[child]
                    root = child
                else:
                    break
            arr[root] = val

        def sift_up(arr, child):
            val = arr[child]
            while child>>1 > 0 and val[1] < arr[child>>1][1]:
                arr[child] = arr[child>>1]
                child >>= 1
            arr[child] = val

        stat = collections.Counter(nums)
        stat = list(stat.items())
        heap = [(0,0)]

        for i in range(k):
            heap.append(stat[i])
            sift_up(heap, len(heap)-1) 

        for i in range(k, len(stat)):
            if stat[i][1] > heap[1][1]:
                heap[1] = stat[i]
                sift_down(heap, 1, k+1) 
        return [item[0] for item in heap[1:]]

leetcode-pp / 91alg-5-daily-check

【Day 20 】2021-09-29 - 347. 前 K 个高频元素 #35

347. 前 K 个高频元素

入选理由

题目地址

前置知识

题目描述

思路

代码

思路

堆排序 + 哈希表

代码

实现语言: C++

复杂度分析

思路

代码

复杂度

思路

Python3代码

复杂度

附录

思路

代码

复杂度：

思路

代码

思路

代码

思路

代码

复杂度

解题思路

思路

thoughts

code

complexity

思路

代码 Python

复杂度分析

思路

代码

思路

代码

复杂度

Idea

Implement (Runtime=O(nlogn), Space=O(n) )

代码

思路

时间&空间

347. Top K Frequent Elements

Algo

Code

Comp

k is the number of n-largest

T: O(Nlogk)

S: O(N+k)

link:

代码 Javascript

复杂度分析

题目

思路

代码

复杂度分析

时间复杂度

空间复杂度

思路

代码

复杂度

题目

思路

代码

复杂度

思路

代码(C++)

复杂度分析

思路：

小顶堆

代码部分：

复杂度：

Time: O(Nlogk)