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
分析：题目很简单，遇到操作符号，连续pop两个数字，计算后push进栈。最后返回栈的唯一一个数字。注意边界条件，如果vector的size为0，返回0。如果在进行数字处理的时候栈为空不能pop数字，返回0。 特别要注意的是，switch与剧中的条件需要能转化为整数，否则会报错。如果不懂的话，就用if吧。哈哈哈，我感觉自己很擅长栈呢（因为比较直观么 ==）。。21ms。
1 class Solution { 2 public: 3 int evalRPN(vector<string>& tokens) { 4 if(tokens.size() == 0) return 0; 5 6 stack<int> result; 7 int ope1 = 0, ope2 = 0; 8 for(int i = 0; i < tokens.size(); i++){ 9 if(tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/") { 10 char temp = tokens[i][0]; 11 if(result.size() == 0) return 0; 12 ope2 = result.top(); 13 result.pop(); 14 15 if(result.size() == 0) return 0; 16 ope1 = result.top(); 17 result.pop(); 18 result.push(calculate(ope1, ope2, temp)); 19 } 20 else result.push(s2int(tokens[i])); 21 } 22 return result.top(); 23 } 24 int s2int(string s){ 25 int negative = 0; 26 if(s[0] == '-') negative = 1; 27 28 int result = 0; 29 int exponential = 1; 30 for(int i = s.length() - 1; i >= negative; i--){ 31 int temp = s[i] - '0'; 32 result += temp * exponential; 33 exponential *= 10; 34 } 35 if(negative) return -result; 36 return result; 37 } 38 int calculate(int ope1, int ope2, char opera){ 39 switch(opera){ 40 case '+': return ope1 + ope2; 41 case '-': return ope1 - ope2; 42 case '*': return ope1 * ope2; 43 case '/': return ope1 / ope2; 44 } 45 } 46 };