代码随想录Day11

 150. 逆波兰表达式求值

#include <vector>
#include <unordered_map>
#include <string>
#include <algorithm>
#include <stack>
#include <queue>using namespace std;class Solution {
public:int evalRPN(vector<string>& tokens) {stack<int> st;for (int i = 0; i < tokens.size(); i++) {if (tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/") {int right = st.top();st.pop();int left = st.top();st.pop();if (tokens[i] == "+") st.push(left + right);if (tokens[i] == "-") st.push(left - right);if (tokens[i] == "*") st.push(left * right);if (tokens[i] == "/") st.push(left / right);}else {st.push(stoi(tokens[i]));}}return st.top();}
};//这题踩的坑：
//left和right反了，因为是用的栈存储，所以先提出来的是右值//要注意的点
//字符串转别的类型是 ato()
//比如转int就是atoi，转long long就是atoll

 239. 滑动窗口最大值

 347.前 K 个高频元素