347. Top K Frequent Elements (sort map)

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.

 Approach #1: C++.

class Solution {
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        unordered_map<int, int> mp;
        for (int i : nums)
            mp[i]++;
        vector<pair<int, int>> v(mp.begin(), mp.end());
        sort(v.begin(), v.end(), cmp);
        vector<int> ans;
        for (int i = 0; i < k; ++i)
            ans.push_back(v[i].first);
        return ans;
    }

private:
    static bool cmp(pair<int, int> a, pair<int, int> b) {
        return a.second > b.second;
    }
};

In order to sort the map with it's value, we can't sort it directly because the iterator type on std::unordered_map is a ForwardIterator, not a RandomAccessIterator, so the first requirement is unsatisfied. The type of the dereferenced iterator is pair<const Key, T>, which is not MoveAssignable (can't assign to const), so the second requirement is also unsatisfied.

we can use a vector to contain the unordered_map, then sort the vector.

Approach #2: Java.

class Solution {
    public List<Integer> topKFrequent(int[] nums, int k) {
        List<Integer>[] bucket = new List[nums.length+1];
        Map<Integer, Integer> frequencyMap = new HashMap<Integer, Integer>();

        for (int n : nums) {
            frequencyMap.put(n, frequencyMap.getOrDefault(n, 0) + 1);
        }

        for (int key : frequencyMap.keySet()) {
            int frequency = frequencyMap.get(key);
            if (bucket[frequency] == null)
                bucket[frequency] = new ArrayList<>();
            bucket[frequency].add(key);
        }

        List<Integer> res = new ArrayList<>();

        for (int pos = bucket.length - 1; pos >= 0 && res.size() < k; --pos) {
            if (bucket[pos] != null)
                res.addAll(bucket[pos]);
        }
        return res;
    }
}

　　

Approach #3: Python.

class Solution(object):
    def topKFrequent(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        """
        return zip(*collections.Counter(nums).most_common(k))[0]

11.Use Counter to extract the top k frequent elements, most_common(k) return a list of tuples, where the first item of the tuple is the element, and the second item of the tuple is the count, Thus,the built-in zip function could be used to extract the first item from the tuples

Time Submitted	Status	Runtime	Language
3 minutes ago	Accepted	40 ms	python
5 minutes ago	Accepted	12 ms	java
20 minutes ago	Accepted	12 ms	cpp