出现次数的TopK问题

#include <unordered_map>
class Solution {
public:
    /**
     * return topK string
     * @param strings string字符串vector strings
     * @param k int整型 the k
     * @return string字符串vector<vector<>>
     */
    vector<vector<string> > topKstrings(vector<string>& strings, int k) {
        // write code here
        vector<vector<string> > ans;
        if (strings.size() == 0 || strings.size() < k)
            return ans;
        // 记录每个元素出现的次数
        unordered_map<string, int> hash;
        for (const auto& string : strings) {
            hash[string]++;
        }
        // 建立小根堆，按照出现次数排序，次数相同按照字典序
        auto cmp = [](const pair<string, int>& p1, const pair<string, int> p2) {
            if (p1.second != p2.second)
                return p1.second > p2.second;
            return p1.first < p2.first;
        };
        // 将全部元素入堆
        priority_queue<pair<string, int>, vector<pair<string, int>>, decltype(cmp)> max_heap(cmp);

        auto p_iter = hash.begin();

        for ( ; p_iter != hash.end(); p_iter++) {
            max_heap.emplace(*p_iter);
            if (max_heap.size() > k)
                max_heap.pop();
        }

        while (!max_heap.empty()) {
            pair<string, int> p = min_heap.top();
            max_heap.pop();
            ans.insert(ans.begin(), vector<string>{p.first, to_string(p.second)} );
        }

        return ans;

    }
};