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
Answer:
对于逆波兰式,一般都是用栈来处理,依次处理字符串,
如果是数值,则push到栈里面
如果是操作符,则从栈中pop出来两个元素,计算出值以后,再push到栈里面,
则最后栈里面剩下的元素即为所求。
class Solution {
public:
int evalRPN (vector<string> &tokens) {
stack<int> operand;
for(int i =0; i< tokens.size(); i++)
{
if ((tokens[i][0] == '-' && tokens[i].size()>1) //negative number
|| (tokens[i][0] >= '0' && tokens[i][0] <= '9')) //positive number
{
operand.push(atoi(tokens[i].c_str()));
continue;
}
int op1 = operand.top();
operand.pop();
int op2 = operand.top();
operand.pop();
if(tokens[i] == "+") operand.push(op2+op1);
if(tokens[i] == "-") operand.push(op2-op1);
if(tokens[i] == "*") operand.push(op2*op1);
if(tokens[i] == "/") operand.push(op2/op1);
}
return operand.top();
}
};
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