LeetCode---Evaluate Reverse Polish Notation

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第二题：Evaluate Reverse Polish Notation

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

这个也是一个很常见的题目，不过在做这道题过程中自己还是忽视了几个问题：

（1）输入的字符串中可能含有负数

（2）输入的字符串中算出来可能出现除数是0的情况

接下来就是采取什么样的数据结构来存了，这个都知道应该是用数据结构里的栈，不过在程序里没有用过Stack，所以自己做这个题的时候还是用了Vector。还可能会有更高效的办法，有时间慢慢研究。

这是我写的这个题目的代码（已AC）：

class Solution {public:int evalRPN(vector<string> &tokens) {if (tokens.size() <= 0){return 0;}vector<int> vecNum;vecNum.clear();for (int i = 0; i < tokens.size(); i++){if (tokens[i].size()> 0 ){if (tokens[i][0] == '-' && tokens[i].size() > 1){int num = 0;for (int j = 1; j < tokens[i].size(); j++){num = (10*num + tokens[i][j] - '0');}num *= -1;vecNum.push_back(num);}else if (tokens[i][0] >= '0' && tokens[i][0] <= '9'){int num = 0;for (int j = 0; j < tokens[i].size(); j++){num = (10*num + tokens[i][j] - '0');}vecNum.push_back(num);}else if(tokens[i].size()==1 && (tokens[i][0] == '+' || tokens[i][0] == '-' || tokens[i][0] == '*' || tokens[i][0] == '/')){switch (tokens[i][0]){case '+':{int firstnum = vecNum[vecNum.size() - 1];vecNum.pop_back();int secondnum = vecNum[vecNum.size() - 1];vecNum.pop_back();vecNum.push_back(firstnum + secondnum);break;}case '-':{int secondnum = vecNum[vecNum.size() - 1];vecNum.pop_back();int firstnum = vecNum[vecNum.size() - 1];vecNum.pop_back();vecNum.push_back(firstnum - secondnum);break;}case '*':{int firstnum = vecNum[vecNum.size() - 1];vecNum.pop_back();int secondnum = vecNum[vecNum.size() - 1];vecNum.pop_back();vecNum.push_back(firstnum * secondnum);break;}case '/':{int secondnum = vecNum[vecNum.size() - 1];vecNum.pop_back();int firstnum = vecNum[vecNum.size() - 1];vecNum.pop_back();if (secondnum == 0){return 0;}vecNum.push_back(firstnum / secondnum);break;}default:break;}}}}int ret = vecNum[0];return ret;}};

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