LeetCode150:Evaluate Reverse Polish Notation

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

解题思路:

很简单的一题,直接利用栈实现,不多说了

实现代码:

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

/*
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
*/ 
class Solution {
public:
    int evalRPN(vector<string> &tokens) {
        stack<string> operand_stack;
        vector<string>::iterator iter;
        for(iter = tokens.begin(); iter != tokens.end(); ++iter)
        {
            if(*iter == "+")//这里就不提取重复代码了,直接简单点 
            {
                string op1 = operand_stack.top();
                operand_stack.pop();
                string op2 = operand_stack.top();
                operand_stack.pop();
                //int ret = atoi(op2.c_str()) + atoi(op1.c_str());
                int ret = stoi(op2) + stoi(op1);//stoi为C++11才有 
                string result = to_string(ret);
                operand_stack.push(result);
                
            }
            else if(*iter == "-")
            {
                string op1 = operand_stack.top();
                operand_stack.pop();
                string op2 = operand_stack.top();
                operand_stack.pop();
                int ret = atoi(op2.c_str()) - atoi(op1.c_str());
                string result = to_string(ret);
                operand_stack.push(result);
                
            }
            else if(*iter == "*")
            {
                string op1 = operand_stack.top();
                operand_stack.pop();
                string op2 = operand_stack.top();
                operand_stack.pop();
                int ret = atoi(op2.c_str()) * atoi(op1.c_str());
                string result = to_string(ret);
                operand_stack.push(result);
                
            }
            else if(*iter == "/")
            {
                string op1 = operand_stack.top();
                operand_stack.pop();
                string op2 = operand_stack.top();
                operand_stack.pop();
                if( atoi(op1.c_str()) == 0)
                    return 0x7FFFFFF;
                int ret = atoi(op2.c_str()) / atoi(op1.c_str());
                string result = to_string(ret);
                operand_stack.push(result);
                
            }
            else
                operand_stack.push(*iter);
        }
        return atoi(operand_stack.top().c_str());
    
    }
 
};

int main(void)
{
    string strs[] = {"4", "13", "5", "/", "+"};
    vector<string> tokens(strs, strs+5);
    Solution solution;
    int ret = solution.evalRPN(tokens);
    cout<<ret<<endl;
    return 0;
}
posted @ 2014-02-16 22:14  mickole  阅读(302)  评论(0编辑  收藏  举报