[LeetCode] Evaluate Reverse Polish Notation

Evaluate Reverse Polish Notation

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

解题思路：

后缀表达式计算的问题，做过计算机的同学应该都知道。遇到数字，进栈；遇到操作符，将前面两个操作数出栈，计算，将结果入栈。最后的结果会在栈中。

代码如下：

class Solution {public:    int evalRPN(vector<string>& tokens) {        stack<int> s;        int len = tokens.size();        int num1, num2;        for(int i = 0; i<len; i++){            if(tokens[i]=="+"){                num1 = s.top();                s.pop();                num2 = s.top();                s.pop();                num1 = num2 + num1;                s.push(num1);            }else if(tokens[i] == "-"){                num1 = s.top();                s.pop();                num2 = s.top();                s.pop();                num1 = num2 - num1;                s.push(num1);            }else if(tokens[i] == "*"){                num1 = s.top();                s.pop();                num2 = s.top();                s.pop();                num1 = num2 * num1;                s.push(num1);            }else if(tokens[i] == "/"){                num1 = s.top();                s.pop();                num2 = s.top();                s.pop();                num1 = num2 / num1;                s.push(num1);            }else{                s.push(stringToInt(tokens[i]));            }        }        return s.top();    }        int stringToInt(const string& s){        int len = s.length();        int result = 0;        int sign = 1;        int i = 0;        if(s[0]=='-'){            sign = -1;            i=1;        }        while(i<len){            result = s[i] - '0' + result * 10;            i++;        }        return sign * result;    }};