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

代码如下

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;
    }
};