Q：Evaluate the value of an arithmetic expression in Reverse
Polish Notation.
Valid operators are+,-,*,/. Each operand may be an integer or another expression.
Some examples: ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6
A：
1 class Solution { 2 public: 3 int evalRPN(vector<string> &tokens) { 4 stack<int> numbers; 5 for(auto token : tokens) 6 { 7 if(token == "+" || token == "-" || token == "*"||token == "/") 8 { 9 int a,b,res; 10 b=numbers.top();numbers.pop(); 11 a=numbers.top();numbers.pop(); 12 if(token == "+") 13 res=a+b; 14 else if(token == "-") 15 res=a-b; 16 else if(token == "*") 17 res=a*b; 18 else 19 res=a/b; 20 numbers.push(res); 21 } 22 else 23 { 24 stringstream ss; 25 ss<<token; 26 int temp; 27 ss>>temp; 28 numbers.push(temp); 29 } 30 } 31 return numbers.top(); 32 } 33 }
涉及知识点
1.stack 的使用
//----------------------------------------- 读取堆栈的栈顶元素 #include <stack> #include <iostream> using namespace std; int main() { // 创建堆栈对象 stack<int> s; // 元素入栈 s.push(3); s.push(19); s.push(23); s.push(36); s.push(50); s.push(4); // 元素依次出栈 while(!s.empty()) { // 打印栈顶元素，打印出：4 50 36 23 19 3 cout << s.top() << endl; // 出栈 s.pop(); } return 0; }
2. stringstream的用法
可以作为将数字和字符串相互转化的工具。
std::string name("12345"); int age = 27; stringstream os; os << name; os >> age; // age = 12345 cout<<age<<endl;
std::string name("12345");
int age = 27;
stringstream os;
os << age;
os >> name;
// name:27
cout<<name<<endl;