https://leetcode.com/problems/evaluate-reverse-polish-notation/
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
代码:
class Solution { public: int evalRPN(vector<string>& tokens) { stack<string> stck; for (auto op : tokens) { if (op == "+") { int operand2 = stoi(stck.top()); stck.pop(); int operand1 = stoi(stck.top()); stck.pop(); int result = operand1 + operand2; stck.push(to_string(result)); } else if (op == "-") { int operand2 = stoi(stck.top()); stck.pop(); int operand1 = stoi(stck.top()); stck.pop(); int result = operand1 - operand2; stck.push(to_string(result)); } else if (op == "*") { int operand2 = stoi(stck.top()); stck.pop(); int operand1 = stoi(stck.top()); stck.pop(); int result = operand1 * operand2; stck.push(to_string(result)); } else if (op == "/") { int operand2 = stoi(stck.top()); stck.pop(); int operand1 = stoi(stck.top()); stck.pop(); int result = operand1 / operand2; stck.push(to_string(result)); } else { stck.push(op); } } int res = stoi(stck.top()); stck.pop(); return res; } };
大一的时候记得讲过这个东西 用到 $stack$ 遇到符号的时候从栈里弹出两个进行运算然后把结果再压进去