leetcode-150. Evaluate Reverse Polish Notation

题目描述

• 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

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思路分析

• 思路：由于逆波兰表达式本身不需要括号来限制哪个运算该先进行，因此可以直接利用栈来模拟计算：遇到操作数直接压栈，碰到操作符直接取栈顶的2个操作数进行计算（注意第一次取出来的是右操作数），然后再把计算结果压栈，如此循环下去。最后栈中剩下的唯一一个元素便是整个表达式的值
• 注意除法的除零保护

AC代码：

class Solution {
public:
    int evalRPN(vector<string> &tokens) {
        int ret=0;
        stack<int> s;  //将tokens往里面丢
        int L=0,R=0;  //左右操作树
        for(int i=0;i<tokens.size();i++)
            {
            if(tokens[i]=="+")
                {
                R=s.top(); //先是右操作数
                s.pop();
                L=s.top();
                s.pop();
                ret=L+R;
                s.push(ret);
            }else if(tokens[i]=="-")
                {
                R=s.top();
                s.pop();
                L=s.top();
                s.pop();
                ret=L-R;
                s.push(ret);
            }else if(tokens[i]=="*")
                {
                R=s.top();
                s.pop();
                L=s.top();
                s.pop();
                ret=L*R;
                s.push(ret);
            }else if(tokens[i]=="/")
                {
                R=s.top();
                s.pop();
                L=s.top();
                s.pop();
                if(R!=0)
                    ret=L/R;
                else
                    return -1;
                s.push(ret);
            }else
                {
                s.push(atoi(tokens[i].c_str()));
            }
        }//end for
         
        return s.top();
    }
};

#include<iostream>
#include<math.h>

#include <vector>
#include<string>
#include<deque>
#include <stack>
#include <queue>
#include<map>
#include <set>

using namespace std;


//evluate reverse polish notation
class Solution {
public:
	int str2digit(string str)
	{


		int ret = 0;

		const char *src = str.c_str();
		ret = atoi(src);

		//for (int i = 0; i < str.size();i++)
		//{
		//	if ()
		//	{
		//	}
		//}
	}

	string digit2str(int data)
	{
		string ret;
		char temp[128] = " ";
		sprintf_s(temp, "%d", data); //./solution.h:25:3: error: use of undeclared identifier 'sprintf_s'

		ret = temp;
		return ret;
	}
	int evalRPN(vector<string> &tokens) {
		stack<string> sta;
		int oper1, oper2,oper3;
		for (unsigned int i = 0; i < tokens.size();i++)
		{
			if (tokens[i] == "+")
			{
				oper2 = str2digit(sta.top());
				sta.pop();
				oper1 = str2digit(sta.top());
				sta.pop();

				oper3 = oper1 + oper2;
				sta.push(digit2str(oper3));

			}else if (tokens[i]=="-")
			{
				oper2 = str2digit(sta.top());
				sta.pop();
				oper1 = str2digit(sta.top());
				sta.pop();

				oper3 = oper1 - oper2;
				sta.push(digit2str(oper3));
			}
			else if (tokens[i]=="*")
			{
				oper2 = str2digit(sta.top());
				sta.pop();
				oper1 = str2digit(sta.top());
				sta.pop();

				oper3 = oper1 * oper2;
				sta.push(digit2str(oper3));
			}
			else if (tokens[i]=="/")
			{
				oper2 = str2digit(sta.top());
				sta.pop();
				oper1 = str2digit(sta.top());
				sta.pop();

				oper3 = oper1 / oper2;  //除零保护
				sta.push(digit2str(oper3));
			}
			else
			{
				sta.push(tokens[i]);
			}
		}
		return str2digit(sta.top());
	}
};


int main()
{

	return 0;
}

reference

• 表达式求值--堆栈应用

• 中缀表达式求值问题