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.

python代码最后用到zip(*a)

假设a1=[1,2,3]

a2=[4,5,6]

c=zip(a1,a2) =[(1,4),(2,5),(3,6)]

zip(*c) 就是解包为a1,a2

class Solution(object):    def topKFrequent(self, nums, k):        """        :type nums: List[int]        :type k: int        :rtype: List[int]        """        frequencymap={};        for i in nums:            if frequencymap.has_key(i):                frequencymap[i] = frequencymap[i] + 1            else:                frequencymap[i] = 0                frequencymap[i] = frequencymap[i] + 1        Sortedmap=sorted(frequencymap.iteritems(),key=lambda d:d[1],reverse=True)        ##l=[]        ##for i in Sortedmap[:k]:        ##    l.append(i[0])                    return zip(*Sortedmap[:k])[0]

vector<int> topKFrequent(vector<int>& nums, int k) {    unordered_map<int, int> m;    for (int num : nums)        ++m[num];          vector<vector<int>> buckets(nums.size() + 1);     for (auto p : m)        buckets[p.second].push_back(p.first);//buckets是一个数组，下标索引是出现的次数，对应的值是出现的这个数              vector<int> ans;    for (int i = buckets.size() - 1; i >= 0 && ans.size() < k; --i) {        for (int num : buckets[i]) {            ans.push_back(num);            if (ans.size() == k)                break;        }    }            return ans;}