【leetcode】Top K Frequent Elements

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.

利用桶排序，分三步解决，时间代价O(n)：

统计每个数的出现频率，存在unordered_map中
定义一个vector<vector<int>>容器，将每个频率对应的数存在vector中
倒序遍历容器，依次将k个数存入结果容器中

class Solution {
public:
vector<int> topKFrequent(vector<int>& nums, int k)
{
unordered_map<int, int> dict;
vector<int> result;
for (auto n : nums)
++dict
;
vector< vector<int> > freq(nums.size() + 1);
for (auto p : dict)
freq[p.second].push_back(p.first);
for (int i = freq.size() - 1; i >= 0 && result.size() < k; --i)
{
for (auto n : freq[i])
{
result.push_back(n);
if (result.size() == k)
break;
}
}
return result;
}
};