leetcode栈--1、evaluate-reverse-polish-notation（逆波兰表达式）

题目描述

 

Evaluate the value of an arithmetic expression inReverse 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：使用栈结构来存储。遇到数字存入栈中，遇到符号，弹出栈顶的两个元素，进行操作，并将结果放入栈中。

class Solution {
public:
    int strtoint(string s) // string转int
    {
        int result = 0;
        int base = 1;//记录属于的位
        int t = 1;//记录正负号
        if(s[0] == '-')
            t = -1;
        for(int i= s.size() - 1;i>=0;i--)
        {
            if(s[i]>='0' && s[i]<='9')//此处一定注意0和9加''
            {
                result += base * (s[i] - '0');
                base *= 10;
            }
        }
        return result*t;
    }
    int evalRPN(vector<string> &tokens) {
        int n = tokens.size();
        if(tokens.empty() || n<=0)
            return 0;
        stack<int> s;
        for(int i=0;i<n;i++)
        {
            if(tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/")
            {
                int num2 = s.top();//先弹出的为右操作数
                s.pop();
                int num1 = s.top();
                s.pop();
                if(tokens[i] == "+")
                {
                    s.push(num1+num2);
                }
                else if(tokens[i] == "-")
                {
                    s.push(num1-num2);
                }
                else if(tokens[i] == "*")
                {
                    s.push(num1*num2);
                }
                else
                {
                    s.push(num1/num2);
                }
            }
            else
            {
                s.push(strtoint(tokens[i]));
            }
        }
        return s.top();
    }
};