150. Evaluate Reverse Polish Not

题目链接
tag:

• Medium；

question：
  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

思路：
  逆波兰表达式就是把操作数放前面，把操作符后置的一种写法，我们通过观察可以发现，第一个出现的运算符，其前面必有两个数字，当这个运算符和之前两个数字完成运算后从原数组中删去，把得到一个新的数字插入到原来的位置，继续做相同运算，直至整个数组变为一个数字。仔细想想，这道题应该是栈的完美应用，从前往后遍历数组，遇到数字则压入栈中，遇到符号，则把栈顶的两个数字拿出来运算，把结果再压入栈中，直到遍历完整个数组，栈顶数字即为最终答案。代码如下:

class Solution {
public:
    int evalRPN(vector<string>& tokens) {
        if (tokens.size() == 1)
            return stoi(tokens[0]);
        stack<int> s;
        for (int i=0; i<tokens.size(); ++i) {
            if (tokens[i] != "+" && tokens[i] != "-" && tokens[i] != "*" && tokens[i] != "/")
                s.push(stoi(tokens[i]));
            else {
                int num1 = s.top();
                s.pop();
                int num2 = s.top();
                s.pop();
                if (tokens[i] == "+") s.push(num2 + num1);
                if (tokens[i] == "-") s.push(num2 - num1);
                if (tokens[i] == "*") s.push(num2 * num1);
                if (tokens[i] == "/") s.push(num2 / num1);
            }
        }
        return s.top();
    }
};