LeetCode刷题笔记（Evaluate Reverse Polish Notation）

刚刚又刷了一道题，感觉难度不是很大，下面就和大家分享一下经验吧！

题目如下：

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

题意分析： 

计算以逆波兰式表示的算术表达式的值。有效的运算符仅包括+, -, *, /，且每个操作数应该是一个整数或者是一个表达式。

注：①两个整数相除，结果向零取整；

       ②被给的逆波兰表达式总是有效的。

解答如下：

方法一（堆栈法）

创建一个存放整型变量的栈，用for循环遍历字符串向量，当遇到字符串数字时，先将其转换成整型数字（用stoi函数），然后在压入堆栈；当遇到字符串运算符时，先弹出最后进栈的两个整型数字，然后用字符串运算符对他们进行计算（注意操作数的顺序），再将计算的结果压入栈中。最后，待循环结束后，返回堆栈中的计算结果即可。

class Solution{
public:
    int evalRPN( vector<string>& tokens ){
        stack<int> stack;    //新建堆栈
        int a;
        int b;
        for (int i = 0; i < tokens.size(); i++) {       //遍历字符串向量，并进行相关操作
            if ( tokens[i] != "+" && tokens[i] != "-" && tokens[i] != "*" && tokens[i] != "/") stack.push( stoi(tokens[i]) );
            else if( tokens[i] != "+" ) { b = stack.top(); stack.pop(); a = stack.top(); stack.pop(); stack.push(a + b);}
            else if( tokens[i] != "-" ) { b = stack.top(); stack.pop(); a = stack.top(); stack.pop(); stack.push(a - b);}
            else if( tokens[i] != "*" ) { b = stack.top(); stack.pop(); a = stack.top(); stack.pop(); stack.push(a * b);}
            else if( tokens[i] != "/" ) { b = stack.top(); stack.pop(); a = stack.top(); stack.pop(); stack.push(a / b);}
        }
        return stack.top();                             //返回结果
    }
};

提交后的结果如下：  

方法二（优化方法一）

① 考虑字符串向量长度为1的情况；

② 简化“弹出最后进栈的两个整型数字的操作”

class Solution {
public:
    int evalRPN(vector<string> &tokens) {
        if ( tokens.size() == 1 ) return stoi(tokens[0]);      //优化一
        stack<int> stack;
        for (int i = 0; i < tokens.size(); i++) {
            if (tokens[i] != "+" && tokens[i] != "-" && tokens[i] != "*" && tokens[i] != "/")
                stack.push(stoi(tokens[i]));
            else {                                             //优化二
                int b = stack.top();                        
                stack.pop();
                int a = stack.top();
                stack.pop();
                if (tokens[i] == "+") stack.push(a + b);
                if (tokens[i] == "-") stack.push(a - b);
                if (tokens[i] == "*") stack.push(a * b);
                if (tokens[i] == "/") stack.push(a / b);
            }
        }
        return stack.top();
    }
};

 提交后的结果如下：  

 日积月累，与君共进，增增小结，未完待续。