2. evaluate-reverse-polish-notation

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

思路：题目意思是给定一个逆波兰表达式，求该表达式的值。

利用栈来模拟计算：遇到操作数直接压栈，碰到操作符取栈顶的2个元素进行计算（第一次取出来的是右操作数），然后再把计算结果压栈，依次循环下去，直到栈中剩下一个元素，就是整个表达式的值。

class Solution {
public:int evalRPN(vector<string> &tokens) {int len = tokens.size();if(len == 0)return 0;if(len == 1)return stoi(tokens[0].c_str());stack<int> stack;for(int i = 0; i < len; ++i){if(tokens[i] != "+" && tokens[i] != "-" && tokens[i] != "*" && tokens[i] != "/")stack.push(stoi(tokens[i].c_str()));else{int tmp = stack.top();stack.pop();if(tokens[i] == "+")tmp += stack.top();if(tokens[i] == "-")tmp = stack.top() - tmp;if(tokens[i] == "*")tmp *= stack.top();if(tokens[i] == "/"){if(tmp == 0)return 0;tmp = stack.top()/tmp;}stack.pop();stack.push(tmp);}}if(stack.size() == 1)return stack.top();elsereturn 0;}
};

stoi(tokens[i].c_str())
/*这里要使用c_str()，因为stoi的参数是const char*，所以在用string时，必须先用c_str()将string转化为char* */

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