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
从stack里面pop出来的时候,谁是num1谁是num2要注意一下顺序。
java的String没法直接比较是不是数字,除非转成char一个一个比,或者用正则表达。
java里面String的比较,应该用s1.equals(s2)而不是直接s1==s2
java里面set的表达方法:
HashSet<String> operator = new HashSet<String>();
operator.add("+");
if(operator.contains("+"))
{...}
class Solution {
public int evalRPN(String[] tokens) {
int cur = 0;
int num1 = 0;
int num2 = 0;
int num3 = 0;
HashSet<String> operator = new HashSet<String>();
operator.add("+");
operator.add("-");
operator.add("/");
operator.add("*");
Stack<Integer> stack = new Stack<Integer>();
for(String c : tokens)
{
if(!operator.contains(c))
stack.add(Integer.parseInt(c));
else
{
num2 = stack.pop();
num1 = stack.pop();
// System.out.println(num1);
// System.out.println(num2);
if(c.equals("/"))
num3 = num1 / num2;
else if(c.equals("+"))
num3 = num1 + num2;
else if(c.equals("-"))
num3 = num1 - num2;
else if(c.equals("*"))
num3 = num1 * num2;
stack.add(num3);
// System.out.println(c);
// System.out.println(num3);
}
}
return stack.pop();
}
}