LeetCode-150-Evaluate Reverse Polish Notation

算法描述：

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

解题思路：逆波特兰表达式，用栈辅助模拟。

    int evalRPN(vector<string>& tokens) {if(tokens.size()==0) return 0;stack<int> stk;for(auto c : tokens){if(c == "+" || c == "-" || c== "*" || c=="/"){int right = stk.top();stk.pop();int left = stk.top();stk.pop();int ans;if(c == "+") ans = left + right;if(c == "-") ans = left - right;if(c == "*") ans = left * right;if(c == "/") ans = left / right;stk.push(ans);}else{stk.push(stoi(c));}}return stk.top();}

转载于:https://www.cnblogs.com/nobodywang/p/10354128.html

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