[LeetCode] Top K Frequent Elements 前K个高频元素

时间:2024-04-23 17:07:36

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.

这道题给了我们一个数组,让统计前k个高频的数字,那么对于这类的统计数字的问题,首先应该考虑用 HashMap 来做,建立数字和其出现次数的映射,然后再按照出现次数进行排序。可以用堆排序来做,使用一个最大堆来按照映射次数从大到小排列,在 C++ 中使用 priority_queue 来实现,默认是最大堆,参见代码如下:

解法一:

class Solution {
public:
vector<int> topKFrequent(vector<int>& nums, int k) {
unordered_map<int, int> m;
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 = ; i < k; ++i) {
res.push_back(q.top().second); q.pop();
}
return res;
}
};

当然,既然可以使用最大堆,还有一种可以自动排序的数据结构 TreeMap,也是可以的,这里就不写了,因为跟上面的写法基本没啥区别,就是换了一个数据结构。这里还可以使用桶排序,在建立好数字和其出现次数的映射后,按照其出现次数将数字放到对应的位置中去,这样从桶的后面向前面遍历,最先得到的就是出现次数最多的数字,找到k个后返回即可,参见代码如下:

解法二:

class Solution {
public:
vector<int> topKFrequent(vector<int>& nums, int k) {
unordered_map<int, int> m;
vector<vector<int>> bucket(nums.size() + );
vector<int> res;
for (auto a : nums) ++m[a];
for (auto it : m) {
bucket[it.second].push_back(it.first);
}
for (int i = nums.size(); i >= ; --i) {
for (int j = ; j < bucket[i].size(); ++j) {
res.push_back(bucket[i][j]);
if (res.size() == k) return res;
}
}
return res;
}
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/347

类似题目:

Kth Largest Element in an Array

Word Frequency

Sort Characters By Frequency

Split Array into Consecutive Subsequences

Top K Frequent Words

K Closest Points to Origin

参考资料:

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

https://leetcode.com/problems/top-k-frequent-elements/discuss/81602/Java-O(n)-Solution-Bucket-Sort

https://leetcode.com/problems/top-k-frequent-elements/discuss/81635/3-Java-Solution-using-Array-MaxHeap-TreeMap

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