题意：写一个计算器，包括加减乘除
Input: s = “3+2*2”
Output: 7
链接
https://leetcode.com/problems/basic-calculator-iii/description/
https://leetcode.com/problems/basic-calculator-ii/description/
需要有一个operator stack以及一个num数字的stack，operator stack中的符号新来的优先级比后来的优先级要小的话那我就要计算
有个取数字的模版要记住：
if(isdigit(s[i])) {int j = i;while(j < s.size() && isdigit(s[j])) j++;int x = stoi(s.substr(i, j-i));num.push(x);i = j - 1;
class Solution {
public:int calculate(string s) {stack<int> num;stack<char> oper;unordered_map<char, int> pr = {{'+', 1}, {'-', 1}, {'*', 2}, {'/', 2}, {'(', 0}};for(int i = 0; i < s.size(); i++) {if(isdigit(s[i])) {int j = i;while(j < s.size() && isdigit(s[j])) j++;int x = stoi(s.substr(i, j-i));num.push(x);i = j - 1;} else if(pr.count(s[i])) {while(oper.size() && pr[s[i]] <= pr[oper.top()]) {get(num, oper);}oper.push(s[i]);} else if(s[i] == ' ') {continue;} else if (s[i] == '(') {oper.push(s[i]);} else if (s[i] == ')') {while(oper.top()!= '(') {get(num, oper);}oper.pop();}}while(oper.size()) {get(num, oper);}return num.top();}void get(stack<int>& num, stack<char>& oper) {int num1 = num.top();num.pop();int num2 = num.top();num.pop();char op = oper.top();oper.pop();int result = 0;switch(op) {case '*': result = num1 * num2; break;case '+': result = num1 + num2; break;case '-': result = num2 - num1; break;case '/': result = num2 / num1; break;}num.push(result);}
};