LeetCode150 Evaluate Reverse Polish Notation 计算逆波兰表达式

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

题源：here；完整实现：here

思路：

最好的思路就是参看维基百科。代码如下：

class Solution {
public:
	int evalRPN(vector<string>& tokens) {
		stack<int> s;
		for (auto it = tokens.begin(); it != tokens.end(); it++) {
			if (isOperand(*it)) {
				int tmp1 = s.top();
				s.pop();
				int tmp2 = s.top();
				s.pop();

				int res = opend(tmp1, *it, tmp2);
				s.push(res);
			}
			else {
				s.push(stoi(*it));
			}
		}
		return s.top();
	}

	bool isOperand(string i) {
		if (i == "+" || i == "-" || i == "*" || i == "/")
			return true;
		else
			return false;
	}

	int opend(int tmp1, string ope, int tmp2) {
		int res = 0;
		switch (ope.c_str()[0]) {
		case '+': {res = tmp2+tmp1; break; }
		case '-': {res = tmp2-tmp1; break; }
		case '*': {res = tmp2*tmp1; break; }
		case '/': {res = tmp2/tmp1; break; }
		default: break;
		}
		return res;
	}
};