[leetcode] 347. Top K Frequent Elements

[leetcode] 347. Top K Frequent Elements

Description

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.

分析

题目的意思是：求一个数组里面前k个频繁元素。

首先建立一个map，然后将map的key value值互换，放进桶里面，然后逆转桶，这样一个一个从桶里面取数据，就是前k个了。

input: nums = [ 1,1,1,1,1 ]map[1] = 5then bucket index range should be 0~5, that why bucket size is nums.size() +1.

代码

class Solution {public: vector topKFrequent(vector& nums, int k) { unordered_map mp; for(auto num:nums){ mp[num]++; } vector> buckets(nums.size()+1); for(auto item:mp){ buckets[item.second].push_back(item.first); } reverse(buckets.begin(),buckets.end()); vector result; for(auto &bucket:buckets){ for(auto i:bucket){ result.push_back(i); if(result.size()==k){ return result; } } } return result; }};

参考文献

​​347. Top K Frequent Elements​​

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