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LeetCode-150-Evaluate Reverse Polish Notation

来源:互联网 收集:自由互联 发布时间:2021-06-23
算法描述: Evaluate the value of an arithmetic expression inReverse Polish Notation. Valid operators are + , - , * , / . Each operand may be an integer or another expression. Note: Division between two integers should truncate toward z

算法描述:

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

解题思路:逆波特兰表达式,用栈辅助模拟。

    int evalRPN(vector<string>& tokens) {
        if(tokens.size()==0) return 0;
        stack<int> stk;
        for(auto c : tokens){
            if(c == "+" || c == "-" || c== "*" || c=="/"){
                int right = stk.top();
                stk.pop();
                int left = stk.top();
                stk.pop();
                int ans;
                if(c == "+") ans = left + right;
                if(c == "-") ans = left - right;
                if(c == "*") ans = left * right;
                if(c == "/") ans = left / right;
                stk.push(ans);
            }else{
                stk.push(stoi(c));
            }
        }
        return stk.top();
    }
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