根据 逆波兰表示法，求表达式的值。
有效的运算符包括 +, -, *, / 。每个运算对象可以是整数，也可以是另一个逆波兰表达式。
说明：
整数除法只保留整数部分。
给定逆波兰表达式总是有效的。换句话说，表达式总会得出有效数值且不存在除数为 0 的情况。
示例 1：
输入: ["2", "1", "+", "3", "*"]
输出: 9
解释: 该算式转化为常见的中缀算术表达式为：((2 + 1) * 3) = 9
示例 2：
输入: ["4", "13", "5", "/", "+"]
输出: 6
解释: 该算式转化为常见的中缀算术表达式为：(4 + (13 / 5)) = 6
示例 3：
输入: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
输出: 22
解释:
该算式转化为常见的中缀算术表达式为：
((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
逆波兰表达式：
逆波兰表达式是一种后缀表达式，所谓后缀就是指算符写在后面。
平常使用的算式则是一种中缀表达式，如 ( 1 + 2 ) * ( 3 + 4 ) 。
该算式的逆波兰表达式写法为 ( ( 1 2 + ) ( 3 4 + ) * ) 。
逆波兰表达式主要有以下两个优点：
去掉括号后表达式无歧义，上式即便写成 1 2 + 3 4 + * 也可以依据次序计算出正确结果。
适合用栈操作运算：遇到数字则入栈；遇到算符则取出栈顶两个数字进行计算，并将结果压入栈中。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/evaluate-reverse-polish-notation
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。This is a medium-difficulty topic on Leetcode. In fact, there is a sentence that is very important: when you encounter a number, it is pushed into the stack; when it encounters an operator, the two numbers on the top of the stack are taken out for calculation, and the result is pushed onto the stack!
This sentence is basically the idea of solving the problem. If I don't read this sentence, it is basically difficult to think of using the stack to solve this problem. This is also the core idea of solving the inverse Polish expression!
answer:
class Solution {
public int evalRPN(String[] tokens) {
Stack<Integer> stack = new Stack();
int a = 0,b = 0,c = 0;
for(int i = 0;i < tokens.length;i++){
switch(tokens[i]){
case "+":
a = stack.pop();
b = stack.pop();
c = a + b;
stack.push(c);
break;
case "-":
a = stack.pop();
b = stack.pop();
c = b - a;
stack.push(c);
break;
case "*":
a = stack.pop();
b = stack.pop();
c = a*b;
stack.push(c);
break;
case "/":
a = stack.pop();
b = stack.pop();
c = b / a;
stack.push(c);
break;
default:
stack.push(Integer.parseInt(tokens[i]));
break;
}
}
return stack.pop();
}
}