前k个高频单词

题目出处：https://leetcode-cn.com/problems/top-k-frequent-words/

方法1：哈希表+排序
思路：用哈希表记录每个单词出现的次数，再按照每个单词出现的次数降序排列(若不同单词出现的次数相同，则按照字符串大小排列)

class Solution {
public:
    vector<string> topKFrequent(vector<string>& words, int k) {
        unordered_map<string, int> mp;
        for(string & s : words){
            ++mp[s]; 
        }
        
        vector<string> ans;
        for(auto & [str, cnt] : mp){
            ans.push_back(str);   
        }

        sort(ans.begin(), ans.end(), [&](const string &a, const string &b){
            if(mp[a] == mp[b]){
                return a < b;
            }
            return mp[a] > mp[b];
        });
        
        ans.erase(ans.begin() + k, ans.end());
        return ans;
    }
};

方法2：优先级队列+哈希表
topK问题：前k个最大元素用最小堆，前k个最小元素用最大堆

typedef pair<string, int> ll;

class Solution {
public:
    vector<string> topKFrequent(vector<string>& words, int k) {
        unordered_map<string, int> mp;
        for(string & s : words){
            ++mp[s];  
        }
        
        auto cmp = [](const ll & a, const ll & b){
            if(a.second == b.second){
                return a < b;
            }
            return a.second > b.second;
        };

        priority_queue<ll, vector<ll>, decltype(cmp)> q(cmp); //构造小根堆

        for(auto & ele : mp){
            q.push(ele);
            if(q.size() > k){
               q.pop();  
            }     
        }
        vector<string> ret(k);
        for (int i = k - 1; i >= 0; i--) {
            ret[i] = q.top().first;
            q.pop();
        }

        return ret;
    }
};

注意点

上述两个代码均用lambda表达式重写priority_queue方法、重写sort方法。
仿函数版本

typedef pair<string, int> ll;

struct cmp{   //仿函数
    bool operator()(const ll &a, const ll &b){
        if(a.second == b.second){
            return a < b;
        }
        return a.second > b.second;
    }
};

class Solution {
public:
    vector<string> topKFrequent(vector<string>& words, int k) {
        unordered_map<string, int> mp;
        for(string & s : words){
            ++mp[s];  
        }

        priority_queue<ll, vector<ll>, cmp> q; //构造小根堆

        for(auto & ele : mp){
            q.push(ele);
            if(q.size() > k){
               q.pop();  
            }     
        }
        
        vector<string> ans(k);
        int index = k-1;
        while(!q.empty()){
            ans[index--] = q.top().first;
            q.pop();  
        }

        return ans;
    }
};