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
此题是求解后缀表达式，题目比较简单，开一个数据stack即可，此题的改进版本是引入括号，此时要开一个数据stack和运算符stack
bool isOperator(string& op){ if(op == "+" || op == "-" || op == "*" || op == "/") return true; else return false; } int calc(int a, int b, char op){ switch(op){ case ‘+‘:return a+b; case ‘-‘:return a-b; case ‘*‘:return a*b; case ‘/‘:return a/b; } } int evalRPN(vector<string> &tokens){ stack<int> num; for(int i = 0 ; i < tokens.size(); ++ i){ if(!isOperator(tokens[i])){ num.push(atoi(tokens[i].c_str())); }else{ int num2 = num.top(); num.pop(); int num1 = num.top(); num.pop(); char op = tokens[i][0]; num.push(calc(num1,num2,op)); } } return num.top(); }
Leetcode Evaluate Reverse Polish Notation,布布扣,bubuko.com
Leetcode Evaluate Reverse Polish Notation
原文地址：http://www.cnblogs.com/xiongqiangcs/p/3794372.html
