题目描述:
Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are +
, -
, *
, /
. Each operand may be an integer or another expression.
Some examples:
["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9 ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6
思路:遍历逆波兰式的每一项。如果是数字的话,则入栈。如果是运算符,则让栈顶的两个元素出栈,做相应的运算后,将运算的结果入栈。遍历结束后,栈中剩余的元素即为逆波兰式的运算结果。
代码:
int evalRPN(vector<string> &tokens)
{
int length = tokens.size();
stack<int> num_stack;
for(int i = 0;i < length;i++)
{
if(strcmp(tokens[i].c_str(),"*") == 0)
{
int number1 = num_stack.top();
num_stack.pop();
int number2 = num_stack.top();
num_stack.pop();
int result = number1 * number2;
num_stack.push(result);
}
else if(strcmp(tokens[i].c_str(),"/") == 0)
{
int number1 = num_stack.top();
num_stack.pop();
int number2 = num_stack.top();
num_stack.pop();
int result = number1 / number2;
num_stack.push(result);
}
else if(strcmp(tokens[i].c_str(),"+") == 0)
{
int number1 = num_stack.top();
num_stack.pop();
int number2 = num_stack.top();
num_stack.pop();
int result = number1 + number2;
num_stack.push(result);
}
else if(strcmp(tokens[i].c_str(),"-") == 0)
{
int number1 = num_stack.top();
num_stack.pop();
int number2 = num_stack.top();
num_stack.pop();
int result = number1 - number2;
num_stack.push(result);
}
else
{
int num = atoi(tokens[i].c_str());
num_stack.push(num);
}
}
return num_stack.top();
}