LeetCode: 150. Evaluate Reverse Polish Notation

题目描述

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

题目大意： 计算逆波兰表达式。

解题思路

依次遍历表达式的每一项，如果是数字，则压入栈内，如果是符号，从栈顶取出两个数来做操作，并将其结果压入栈顶。

AC 代码

class Solution {
public:
    int evalRPN(vector<string>& tokens) {
        stack<int> numStk;
        for(string token : tokens)
        {
            if(token != "+" && token != "-" && token != "*" && token != "/")
            {
                numStk.push(stoi(token));
                continue;
            }

            int rhv = numStk.top();
            numStk.pop();
            int lhv = numStk.top();
            numStk.pop();

            if(token  == "+")
            {
                numStk.push(lhv+rhv);
            }
            else if(token == "-")
            {
                numStk.push(lhv-rhv);
            }
            else if(token == "*")
            {
                numStk.push(lhv*rhv);
            }
            else if(token == "/")
            {
                numStk.push(lhv/rhv);
            }
        }
        return numStk.top();
    }
};