题目
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.
复杂度分析
优先队列->构造最大堆:O(nlgn)
std::map 按照key排序:STL中的map是红黑树(待学习)
19 ms submission
class Solution {
public:
vector<int> topKFrequent(vector<int>& nums, int k) {
//思路其实蛮简单的,就是按出现的频次排一个顺序;
unordered_map<int, int> m;
//自动按照pair[0]进行最大优先队列构造
priority_queue<pair<int, int>> q;
vector<int> res;
for (auto a : nums) ++m[a];
for (auto it : m) q.push({it.second, it.first});
for (int i = 0; i < k; ++i) {
res.push_back(q.top().second); q.pop();
}
return res;
}
}
自己的29ms 找差距
class Solution {
public:
/*
1. 数组已经排好序么?
2. 数组中数字的范围已知么? 遍历map进行k-v存储
3. 输出的数组需要按频率排序么?
*/
vector<int> topKFrequent(vector<int>& nums, int k) {
int len = nums.size();
map<int,int> fmap;
for(int i=0;i<len;++i){
if(fmap.find(nums[i])==fmap.end()){
fmap[nums[i]] = 0;
}else{
fmap[nums[i]]++;
}
}
map<int,vector<int>> bucket;
for(auto it=fmap.begin();it!=fmap.end();++it){
bucket[it->second].push_back(it->first);
}
vector<int> res;
//map会自动根据key值进行排序,所以根据迭代器从后往前找topk就可以。
//用rbegin和rend进行倒序输入
for(auto it=bucket.rbegin();k>0 && it!=bucket.rend();++it){
vector<int> temp = it->second;
while(k>=0 && !temp.empty()){
res.push_back(temp.back());
temp.pop_back();
--k;
}
}
return res;
}
};