题目

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

原题地址

解题思路

代码例如以下

class Solution {

public:

    int evalRPN(vector<string> &tokens) {

        int ret=0;

        int n = tokens.size();

        if(n<=0) return ret;

        stack<int> s;

        int i=0;

        while(i<n){

            string temp = tokens[i];

            int num = atoi(temp.c_str());

            //防止非法输入

            if(num!=0 || (num==0 && temp[0]=='0')){

                s.push(num);

            }else{

                ret = cal(s, tokens[i][0]);

                if(ret == -1) return 0;

            }

            ++i;

        }

        if(!s.empty()) return s.top();

        else return 0;

    }

    int cal(stack<int> &s, char oper){

        if(s.size()<2) return -1;

        int op1 = s.top(); s.pop();

        int op2 = s.top(); s.pop();

        if(oper == '+'){

            s.push(op1+op2);

        }else if(oper == '-'){

            s.push(op2-op1);

        }else if(oper == '*'){

            s.push(op2 * op1);

        }else if(oper == '/'){

            if(op1 == 0) return -1;

            s.push(op2 / op1);

        }else return -1;

        return 0;

    }

};