【LC刷题】DAY10：150 239 347

文章目录

【LC刷题】DAY10：150 239 347
- 150. 逆波兰表达式求值 [link](https://leetcode.cn/problems/evaluate-reverse-polish-notation/description/)
- 239. 滑动窗口最大值 [link](https://leetcode.cn/problems/sliding-window-maximum/description/)
- 347. 前 K 个高频元素

150. 逆波兰表达式求值 link

class Solution {
public:int evalRPN(vector<string>& tokens) {stack<int> stk;int n = tokens.size();for(int i = 0; i< n ; i++){string &token = tokens[i];if(isNumber(token)){stk.push(atoi(token.c_str()));}else{int num2 = stk.top();stk.pop();int num1 = stk.top();stk.pop();switch(token[0]){case '+':stk.push(num1 + num2);break;case '-':stk.push(num1 - num2);break;case '*':stk.push(num1 * num2);break;case '/':stk.push(num1 / num2);break;}}}return stk.top();}bool isNumber(string &token){return!(token == "+" || token == "-" || token == "*" || token == "/");}
};

239. 滑动窗口最大值 link

class Solution {
public:vector<int> maxSlidingWindow(vector<int>& nums, int k) {int n = nums.size();priority_queue<pair<int, int>> q;for(int i = 0; i < k; i++){q.emplace(nums[i], i);}vector<int> ans = {q.top().first};for(int i = k; i < n; i++){q.emplace(nums[i], i);while(q.top().second <= i - k){q.pop();}ans.push_back(q.top().first);}return ans;}
};

347. 前 K 个高频元素

class Solution {
public:static bool cmp(pair<int, int>& m, pair<int, int>& n) {return m.second > n.second;}vector<int> topKFrequent(vector<int>& nums, int k) {unordered_map<int, int> occurrences;for (auto& v : nums) {occurrences[v]++;}// pair 的第一个元素代表数组的值，第二个元素代表了该值出现的次数priority_queue<pair<int, int>, vector<pair<int, int>>, decltype(&cmp)> q(cmp);for (auto& [num, count] : occurrences) {if (q.size() == k) {if (q.top().second < count) {q.pop();q.emplace(num, count);}} else {q.emplace(num, count);}}vector<int> ret;while (!q.empty()) {ret.emplace_back(q.top().first);q.pop();}return ret;}
};