Leetcode [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
逆波兰表达式求值
根据逆波兰表示法，求表达式的值。
有效的运算符包括 +, -, *, / 。每个运算对象可以是整数，也可以是另一个逆波兰表达式。
说明：
整数除法只保留整数部分。
给定逆波兰表达式总是有效的。换句话说，表达式总会得出有效数值且不存在除数为 0 的情况。
示例 1：
输入: [“2”, “1”, “+”, “3”, “*”]
输出: 9
解释: ((2 + 1) * 3) = 9
示例 2：
输入: [“4”, “13”, “5”, “/”, “+”]
输出: 6
解释: (4 + (13 / 5)) = 6
示例 3：
输入: [“10”, “6”, “9”, “3”, “+”, “-11”, “", “/”, "”, “17”, “+”, “5”, “+”]
输出: 22
解释:
((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
这题虽然是Medium水平，但很简单的，easy系列水平。
本来是想用switch case语句 case里不能是string ;所以此题用的if ==
因为定义的栈是整形，所以不能直接入栈。借助stoic()。
class Solution {
public:
int evalRPN( vector& tk ) {
int sum = 0;
stack < int > st ;
for ( int i = 0 ; i < tk.size() ; i++){
if (tk[i] == "+"||tk[i] == "-"||tk[i] == "*"||tk[i] == "/")
{
int s1 = st.top() ;
st.pop() ;
int s2 = st.top();
st.pop() ;
if(tk[i] == "+" )
st.push( s1 + s2 );
else if (tk[i] == "-" )
st.push( s2 - s1 );
else if (tk[i] == "*" )
st.push( s1 * s2 );
else if (tk[i] == "/" )
st.push( s2 / s1 );
}
else st.push ( stoi(tk[i]) );
}
return st.top() ;
}
};