150. Evaluate Reverse Polish Notation
Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are +, -, *, and /. Each operand may be an integer or another expression.
Note that division between two integers should truncate toward zero.
It is guaranteed that the given RPN expression is always valid. That means the expression would always evaluate to a result, and there will not be any division by zero operation.
Example 1:
Input: tokens = ["2","1","+","3","*"]
Output: 9
Explanation: ((2 + 1) * 3) = 9
Example 2:
Input: tokens = ["4","13","5","/","+"]
Output: 6
Explanation: (4 + (13 / 5)) = 6
Example 3:
Input: tokens = ["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
Constraints:
- 1 <= tokens.length <= 104
- tokens[i] is either an operator: "+", "-", "*", or "/", or an integer in the range [-200, 200].
class Solution {
public int evalRPN(String[] tokens) {
Stack<Integer> stack = new Stack<>();
for(int i=0;i<tokens.length;i++){
if(tokens[i].equals("+")){
stack.push(stack.pop() + stack.pop());
}else if(tokens[i].equals("-")){
stack.push(-stack.pop() + stack.pop());
}else if(tokens[i].equals("*")){
stack.push(stack.pop()*stack.pop());
}else if(tokens[i].equals("/")){
int temp= stack.pop();
stack.push(stack.pop()/temp);
}else{
stack.push(Integer.parseInt(tokens[i]));
}
}
return stack.pop();
}
}使用栈的方式求解，注意以下几点
- 减法和除法的计算
- 非符号时如何存入数字