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.

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Solution:

1.使用hashtable獲取每個元素出現的次數

2.使用小根堆heap排序（priority queue實現），獲取前K個出現次數最多的元素pair

3.輸出，注意是小根堆，要reverse一下

時間復雜度：O(N * logK)

1 class CompareDist 2 { 3 public: 4 bool operator()(pair<int, int> n1, pair<int, int> n2) 5 { 6 return n1.second > n2.second; 7 } 8 }; 9 10 class Solution { 11 public: 12 vector<int> topKFrequent(vector<int>& nums, int k) 13 { 14 vector<int> ret; 15 unordered_map<int, int> htable; 16 17 for (int key : nums) // get each key appears times 18 htable[key]++; 19 20 priority_queue<pair<int, int>, vector<pair<int, int> >, CompareDist> sheap; // use min heap to get k biggest 21 for (auto elem : htable) 22 { 23 if (sheap.size() < k) 24 { 25 sheap.push(elem); 26 } 27 else 28 { 29 pair<int, int> heap_min = sheap.top(); 30 if (elem.second > heap_min.second) 31 { 32 sheap.pop(); 33 sheap.push(elem); 34 } 35 } 36 } 37 38 while (!sheap.empty()) 39 { 40 pair<int, int> heap_min = sheap.top(); 41 ret.push_back(heap_min.first); 42 sheap.pop(); 43 } 44 reverse(ret.begin(), ret.end()); 45 46 return ret; 47 } 48 };

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或者直接使用make_heap操作，時間復雜度O(N + KlogN) = O(N)

1 vector<int> topKFrequent(vector<int>& nums, int k) 2 { 3 unordered_map<int, int> htable; 4 for (auto &n : nums) 5 htable[n]++; 6 7 vector<pair<int, int>> heap; 8 for (auto &i : htable) 9 heap.push_back({i.second, i.first}); 10 11 vector<int> result; 12 make_heap(heap.begin(), heap.end()); 13 while (k--) 14 { 15 result.push_back(heap.front().second); 16 pop_heap(heap.begin(), heap.end()); 17 heap.pop_back(); 18 } 19 return result; 20 }

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轉載于:https://www.cnblogs.com/ym65536/p/5537052.html

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