LeetCode(150. Evaluate Reverse Polish Notation)

简介

Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are +, -, , /. Each operand may be an integer or another expression.
Note:
Division between two integers should truncate toward zero.
The given RPN expression is always valid. That means the expression would always evaluate to
a result and there won’t be any divide by zero operation.
Example 1:
Input: [“2”, “1”, “+”, “3”, ""]
Output: 9
Explanation: ((2 + 1) * 3) = 9
Example 2:
Input: [“4”, “13”, “5”, “/”, “+”]
Output: 6
Explanation: (4 + (13 / 5)) = 6
Example 3:
Input: [“10”, “6”, “9”, “3”, “+”, “-11”, “", “/”, "”, “17”, “+”, “5”, “+”]
Output: 22
Explanation:
((10 * (6 / ((9 + 3) * -11))) + 17) + 5
= ((10 * (6 / (12 * -11))) + 17) + 5
= ((10 * (6 / -132)) + 17) + 5
= ((10 * 0) + 17) + 5
= (0 + 17) + 5
= 17 + 5
= 22

解题思路

这是逆波兰式（后缀表达式）的计算。
判断若为数字，将其压入数字栈，若为运算符，则把栈顶两个数字压出，进行运算后，再把新计算的数压入栈。最后栈内只剩一个数，即为运算结果。

代码

#include <iostream>
#include<string>
#include<vector>
#include<stack>
#include <sstream>
using namespace std;

class Solution {
public:
    int evalRPN(vector<string>& tokens) {
        int temp;
        int a, b;
        for(vector<string>::iterator it = tokens.begin(); it != tokens.end(); it++)
        {
            if(*it == "+" || *it == "-" || *it == "*" || *it == "/" && num.empty() != true)
            {
                a = num.top();
                num.pop();
                b = num.top();
                num.pop();
                if(*it == "+")
                {
                    temp = a + b;
                    num.push(temp);
                }
                else if(*it == "-")
                {
                    temp = b - a;
                    num.push(temp);
                }
                 else if(*it == "*")
                {
                    temp = b * a;
                    num.push(temp);
                }
                else
                {
                    temp = b / a;
                    num.push(temp);
                }
            }
            else
            {
                int n;
                n = StrToInt(*it);
                num.push(n);
            }
        }
        return num.top();

    }
private:
    stack<int> num;
    int StrToInt(string str)   //字符串转换为整型
    {
        stringstream ss;
        ss<<str;
        int i;
        ss>>i;
        return i;
    }
};


int main()
{
    vector<string> str;
    //string i;
    str.push_back("10");
    str.push_back("6");
    str.push_back("9");
    str.push_back("3");
    str.push_back("+");
    str.push_back("-11");
    str.push_back("*");
    str.push_back("/");
    str.push_back("17");
    str.push_back("+");
    str.push_back("5");
    str.push_back("+");
    Solution su;
    cout << su.evalRPN(str);
    return 0;
}