【刷题-LeetCode】150 Evaluate Reverse Polish Notation

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150. 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.

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

解 逆波兰表达式中，两个操作数紧跟一个运算符

class Solution {
public:
    int evalRPN(vector<string>& tokens) {
        stack<int>s1;
        for(int i = 0; i < tokens.size(); ++i){
            if(tokens[i].size() == 1 && !isdigit(tokens[i][0])){
                int x1 = s1.top();
                s1.pop();
                int x2 = s1.top();
                s1.pop();
                s1.push(compute(x2, x1, tokens[i]));
            }else{
                s1.push(stoi(tokens[i]));
            }
        }
        return s1.top();
    }
    int compute(int &x1, int &x2, string &op){
        if(op == "+"){
            return x1 + x2;
        }else if(op == "-"){
            return x1 - x2;
        }else if(op == "*"){
            return x1 * x2;
        }else if(op == "/"){
            return x1 / x2;
        }
        return -1;
    }
};

Note : 影响代码速度

1. for循环中直接枚举会慢一些
2. 函数传参时，传值会慢一些，尽量使用传引用

【刷题-LeetCode】150 Evaluate Reverse Polish Notation

标签：run   运算符   ++   ret   visio   public   ref   ide   tor   

原文地址：https://www.cnblogs.com/vinnson/p/13260010.html