[leetcode]Evaluate Reverse Polish Notation

逆波兰表达式，情况不是很复杂，用一个栈就解决了。

#include <stack>
#include <string>
using namespace std;

class Solution {
public:
    int evalRPN(vector<string> &tokens) {
        for (int i = 0; i < tokens.size(); i++) {
            string &token = tokens[i];
            if (token == "+") {
                int a = num_stack.top();
                num_stack.pop();
                int b = num_stack.top();
                num_stack.pop();
                int r = a + b;
                num_stack.push(r);
            } else if (token == "-") {
                int a = num_stack.top();
                num_stack.pop();
                int b = num_stack.top();
                num_stack.pop();
                int r = b - a;
                num_stack.push(r);
            } else if (token == "*") {
                int a = num_stack.top();
                num_stack.pop();
                int b = num_stack.top();
                num_stack.pop();
                int r = a * b;
                num_stack.push(r);
            } else if (token == "/") {
                int a = num_stack.top();
                num_stack.pop();
                int b = num_stack.top();
                num_stack.pop();
                int r = b / a;
                num_stack.push(r);
            } else { // number
                bool neg = (token[0] == '-');
                int pos = 0;
                if (neg) {
                    pos = 1;
                }
                int r = 0;
                for (int k = pos; k < token.length(); k++) {
                    r = r * 10 + (token[k] - '0');
                }
                if (neg) {
                    r = -r;
                }
                num_stack.push(r);
            }
        }
        int result = num_stack.top();
        num_stack.pop();
        return result;
    }

private:
    stack<int> num_stack;
};

中间那些部分可以简化出来。

int o1, o2;  
o2 = numeric.top();  
numeric.pop();  
o1 = numeric.top();  
numeric.pop();  

switch(t[0])
{
    case '+':
        numeric.push(o1 + o2);
        break;
    case '-':
        numeric.push(o1 - o2);
        break;
    case '*':
        numeric.push(o1 * o2);
        break;
    case '/':
        numeric.push(o1 / o2);
        break;
}

第二刷，Annie的解法比较简洁。且其中stoi可以直接将字符串转成整数。

python3，要注意//代表floor division

class Solution:
    def evalRPN(self, tokens: List[str]) -> int:
        stack = []
        for token in tokens:
            if token in ['+', '-', '*', '/']:
                num2 = stack.pop()
                num1 = stack.pop()
                if token == '+':
                    stack.append(num1 + num2)
                elif token == '-':
                    stack.append(num1 - num2)
                elif token == '*':
                    stack.append(num1 * num2)
                elif token == '/':
                    stack.append(int(num1 / num2))
            else:
                stack.append(int(token))
        
        return stack.pop()