LeetCode 150. Evaluate Reverse Polish Notation
2017-05-24 01:45
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题目
题意
分析
代码
Valid operators are
Some examples:
合法的操作符是
下面是一些例子:
逆波兰表达式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:
20 / 20 test cases passed.
Runtime: 9 ms
题意
分析
代码
题目
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
题意
计算出逆波兰表达式(Reverse Polish Notation)的值。合法的操作符是
+,
-,
*,
/。每一个操作数可以是整数或者另一种表达式。
下面是一些例子:
["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9 ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6
分析
逆波兰表达式又叫做后序表达式,我们常看到的表达式为中序表达式,比如a+b,表示为
ab+。人的大脑适合理解中序表达式,但是计算机比较倾向于逆波兰表达式。
逆波兰表达式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,则进栈, 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。
代码
int evalRPN(vector<string>& tokens) { stack<int> myStack; int temp; for(int i=0;i<tokens.size();i++){ if(tokens[i] == "+" || tokens[i] == "-" || tokens[i] == "*" || tokens[i] == "/"){ temp = myStack.top(); myStack.pop(); if(tokens[i] == "+") temp += myStack.top(); else if(tokens[i] == "-") temp = myStack.top() - temp; else if(tokens[i] == "*") temp *= myStack.top(); else if(tokens[i] == "/") temp = myStack.top() / temp; myStack.pop(); myStack.push(temp); } else myStack.push(stoi(tokens[i].c_str())); } return myStack.top(); }
20 / 20 test cases passed.
Runtime: 9 ms
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