[LeetCode] 347. Top K Frequent Elements

轉自LeetCode

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.

<Solution>

想法如下

• 用 unordered_map 來做 hash table 建表，key 是數值，value是該數值的頻率

• 用 priority_queue 根據頻率來做 max heap，這樣只要陸續從 queue 前面拿 k 個，就是答案

code 如下
c++

	class Solution {
	public:
	struct myCmp {
	bool operator() (pair<int, int> a, pair<int, int> b) {
	return a.second < b.second;
	}
	};
	vector<int> topKFrequent(vector<int>& nums, int k) {
	unordered_map<int, int> hashTable;
	priority_queue<pair<int, int>, vector<pair<int,int>>, myCmp> maxHeap;
	//> create hash table
	for(auto n : nums) {
	++hashTable[n];
	}
	//> create max heap
	for(auto h : hashTable) {
	maxHeap.push(h);
	}
	//> find top k frequent
	vector<int> ans(k,0);
	for(int i = 0; i < k; i++) {
	ans[i] = maxHeap.top().first;
	maxHeap.pop();
	}
	return ans;
	}
	};

view raw topKFrequent.cpp hosted with ❤ by GitHub

kotlin 可以更簡化一點

	class Solution {
	fun topKFrequent(nums: IntArray, k: Int): IntArray {
	val map = mutableMapOf<Int,Int>()
	for(n in nums) {
	map[n] = map[n]?.let{it+1} ?: 1
	}
	return map.entries
	.sortedByDescending {it.value}
	.map {it.key}
	.take(k)
	.toIntArray()
	}
	}

view raw top_k_frequent_elements.kt hosted with ❤ by GitHub