算法描述: Evaluate the value of an arithmetic expression inReverse Polish Notation. Valid operators are + , - , * , / . Each operand may be an integer or another expression. Note: Division between two integers should truncate toward z
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(); }