evaluate-reverse-polish-notation
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题目描述
Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are+,-,*,/. Each operand may be an integer or another expression.
Some examples:
["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9 ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6
(a+b)*c=》
ab+c*
那么ab+c*的执行结果如下:
a入栈(0位置); b入栈(1位置); 遇到运算符“+”,将a和b出栈,执行a+b的操作,得到结果d=a+b,再将d入栈(0位置); c入栈(1位置); 遇到运算符“*”,将d和c出栈,执行d*c的操作,得到结果e,再将e入栈(0位置)。
class Solution {public: int str2int(string s) // string转int { int result = 0; int base = 1; int t = 1; // 正负号 if (s[0] == '-') t = -1; for (int i = s.size() - 1; i >= 0; --i) { if (s[i] >= '0' && s[i] <= '9') { result += base * (s[i] - '0'); base *= 10; } } return result * t; } int evalRPN(vector<string> &tokens) { stack<int> k; for (int i = 0; i < tokens.size(); ++i) { if (tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/") { int Num2 = k.top(); // 第一个取出的是右操作数 k.pop(); int Num1 = k.top(); // 左操作数 k.pop(); if (tokens[i] == "+"){ k.push(Num1 + Num2); } else if (tokens[i] == "-"){ k.push(Num1 - Num2); } else if (tokens[i] == "*"){ k.push(Num1 * Num2); } else if (tokens[i] == "/"){ k.push(Num1 / Num2); } } else k.push(str2int(tokens[i])); } return k.top(); // 最后栈剩下一个元素,就是结果 } };
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