题解 | #矩阵的最小路径和#

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<>>
     */
    // 自定义排序规则，先按照second从大到小排序、如果second相等的情况下，按照first从小到大排序
    struct cmp{
            bool operator() (const pair<string, int>& p1, const pair<string, int>& p2) {
                return (p1.second < p2.second) || (p1.second == p2.second && p1.first > p2.first);
            }
        };
    vector<vector<string> > topKstrings(vector<string>& strings, int k) {
        // write code here
        int n = strings.size();
        vector<vector<string> > res;

        unordered_map<string, int> mp;
        for(auto s:strings)
            mp[s]++;

        priority_queue<pair<string, int>, vector<pair<string, int>>, cmp> q;
        int i = 0;
        // 将所有的数据排序
        for(auto m:mp)
            q.push(m);
        // 输出前k个数值
        while(i < k) {
            res.push_back({q.top().first, to_string(q.top().second)});
            q.pop();
            i++;
        }
        return res;
    }
};