LeetCode : 150. Evaluate Reverse Polish Notation

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题目

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

就是简单的栈使用，其中需要注意的点就是，字符串怎么判断是不是数字，和将结果也压入栈中

方法一： .charAt(i),判断字符串的每一个字符是不是数字
方法二：Integer.parseInt(tokens[i]) 直接强制字符串转数字
方法三： Integer.valueOf(tokens[i]),也是直接转
然后用也可以尝试：try catch

try{int num = Integer.parseInt(tokens[i]);stack.add(num);}catch (Exception e) {int b = stack.pop();int a = stack.pop();stack.add(get(a, b, tokens[i]));}

解决代码

class Solution {public int evalRPN(String[] tokens) {Stack<Integer> stack = new Stack<Integer>();int len = tokens.length;int ans = 0;for(int i = 0;i < len; i++){switch (tokens[i]){case "+": ans = Integer.valueOf(stack.pop()) + Integer.valueOf(stack.pop());stack.push(ans);break;case "-":{int temp = Integer.valueOf(stack.pop());ans = (Integer.valueOf(stack.pop()-temp));stack.push(ans);}break;case "*": ans = (Integer.valueOf(stack.pop()) * Integer.valueOf(stack.pop()));stack.push(ans);break;case "/":{int temp = Integer.valueOf(stack.pop());ans = (Integer.valueOf(stack.pop()/temp));stack.push(ans);}break;default: stack.push(Integer.valueOf(tokens[i]));break;}}return stack.pop();     }
}