Evaluate Reverse Polish Notation（堆栈）

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

分析：

/*-----Reverse Polish Notation（逆波兰表达式），又叫做后缀表达式。在通常的表达式中，二元运算符总是置于与之相关的两个运算对象之间，这种表示法也称为中缀表示。波兰逻辑学家J.Lukasiewicz于1929年提出了另一种表示表达式的方法，按此方法，每一运算符都置于其运算对象之后，故称为后缀表示。

优势

它的优势在于只用两种简单操作，入栈和出栈就可以搞定任何普通表达式的运算。其运算方式如下：

如果当前字符为变量或者为数字，则压栈，如果是运算符，则将栈顶两个元素弹出作相应运算，结果再入栈，最后当表达式扫描完后，栈里的就是结果。-----*/

 

思路：利用堆栈来解决这道题。注意，我们还需要实现整数与字符串之间的相互转换。

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

class Solution {  public:      int evalRPN(vector<string> &tokens) {          stack<int> cache;                      for(int i = 0 ; i < tokens.size(); i++){              if(tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/"){                  int num2 = cache.top();                  cache.pop();                  int num1 = cache.top();                  cache.pop();                  cache.push(calculate(num1, num2, tokens[i]));              }              else{                  cache.push(str2int(tokens[i]));              }          }                      return cache.top();      }              int str2int(string s){          int result=0;          int base=1;          for(int i = s.size()-1;i>=0;i--){              if(s[i] == '-' && i == 0){                  result *= -1;              }              else if(s[i] >= '0' && s[i] <= '9'){                  result += base * (s[i] - '0');                  base *= 10;              }          }          return result;      }              int calculate(int num1, int num2, string op){          if(op == "+"){              return num1 + num2;          }          else if(op == "-"){              return num1 - num2;          }          else if(op == "*"){              return num1 * num2;          }else if(op == "/"){              return num1 / num2;          }      }  };

　其他方法：

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

class Solution { public:     int evalRPN(vector<string>& tokens) {         stack<int> s;         for (auto t : tokens) {                                         //自动类型推断             if (t == "+" || t == "-" || t == "*" || t == "/") {                 int y = s.top(); s.pop();                 int x = s.top(); s.pop();                 int z = 0;                 switch (t.front()) {                     case '+' :                         z = x + y;                         break;                     case '-' :                         z = x - y;                         break;                     case '*' :                         z = x * y;                         break;                     case '/' :                         z = x / y;                         break;                 }                 s.push(z);             } else {                 s.push(stoi(t));     //字符串怎么转数值用函数 std::stoi（）函数原型：                                          //int stoi (const string& str, size_t* idx = 0, int base = 10); base 是进制                                                             }         }         return s.top();     } };

　　使用is_operator更简洁：

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

class Solution { public:     int evalRPN(vector<string>& tokens) {     stack<int> stn;     for(auto s:tokens) {         if(s.size()>1 || isdigit(s[0])) stn.push(stoi(s));         else {             auto x2=stn.top(); stn.pop();             auto x1=stn.top(); stn.pop();             switch(s[0]) {                 case '+': x1+=x2; break;                 case '-': x1-=x2; break;                 case '*': x1*=x2; break;                 case '/': x1/=x2; break;             }             stn.push(x1);         }     }     return stn.top(); } };