计算逆波兰式 Evaluate Reverse Polish Notation

问题： 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

.思路：问题只有二元操作符，所以所有计算都是先出现左右操作数，再出现操作符。 用栈读取：见到数字就吞入栈中；见到操作符就把栈中的两个数吐出，计算之后再吞入栈中。

class Solution {
	public:
		int evalRPN(vector &tokens) {
			stack s;
			int left, right, sum;
			for(int i=0;i