150. Evaluate Reverse Polish Notation

problem description

Evaluate the value of an arithmetic expression in Reverse Polish Notation.

Valid operators are +, -, *, /. Each operand may be an integer or another expression.

Note:

Division between two integers should truncate toward zero. The given RPN expression is always valid. That means the expression would always evaluate to a result and there won't be any divide by zero operation.

Example 1:

Input: ["2", "1", "+", "3", "*"]
Output: 9
Explanation: ((2 + 1) * 3) = 9

Example 2:

Input: ["4", "13", "5", "/", "+"]
Output: 6
Explanation: (4 + (13 / 5)) = 6

Example 3:

Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
Output: 22
Explanation:
  ((10 * (6 / ((9 + 3) * -11))) + 17) + 5
= ((10 * (6 / (12 * -11))) + 17) + 5
= ((10 * (6 / -132)) + 17) + 5
= ((10 * 0) + 17) + 5
= (0 + 17) + 5
= 17 + 5
= 22

problem description

波兰表达式，直接使用一个栈解决，碰到数字压入，碰到符号就计算，并且将计算结果压入栈中

code

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 y = st.top();
                st.pop();
                int x = st.top();
                st.pop();
                if (tokens[i] == "+") {
                    st.push(x+y);
                }
                if (tokens[i] == "-") {
                    st.push(x-y);
                }
                if (tokens[i] == "*") {
                    st.push(x*y);
                }
                if (tokens[i] == "/") {
                    st.push(x/y);
                }
            } else {
                st.push(stoi(tokens[i]));
            }
        }
        return st.top();
    }
};

algorithm analysis

程序只需要线性时间遍历数组即可，时间复杂度O(n)