leetcode(java) 150. Evaluate Reverse Polish Notation-优快云博客

本文链接：https://blog.youkuaiyun.com/jmqxnxg/article/details/128686219

该问题使用栈来解决，遍历输入的tokens数组，遇到数字时压入栈，遇到运算符则从栈中弹出两个数字进行运算，结果再入栈。最后栈顶的数值即为表达式的值。

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Q: You are given an array of strings tokens that represents an arithmetic expression in a Reverse Polish Notation.

Evaluate the expression. Return an integer that represents the value of the expression.

Note that:

    The valid operators are '+', '-', '*', and '/'.
    Each operand may be an integer or another expression.
    The division between two integers always truncates toward zero.
    There will not be any division by zero.
    The input represents a valid arithmetic expression in a reverse polish notation.
    The answer and all the intermediate calculations can be represented in a 32-bit integer.

A: I can use a stack to solve this problem. When the element is a number, I will push it into the stack. When the element is an operator(+-*/), I will pop two number from stack and calculate it, push result into stack.

 public int evalRPN(String[] tokens) {
        Stack stack = new Stack();
        int a, b;
        for(String c: tokens){
            switch(c){
                case "+":   
                            a = (int)stack.pop();
                            b = (int)stack.pop();
                            stack.push(a+b);
                            break;
                case "-":   
                            a = (int)stack.pop();
                            b = (int)stack.pop();
                            stack.push(b-a);
                            break;
                case "*":   
                            a = (int)stack.pop();
                            b = (int)stack.pop();
                            stack.push(a*b);
                            break;
                case "/":   
                            a = (int)stack.pop();
                            b = (int)stack.pop();
                            stack.push(b/a);
                            break;
                default:    stack.push(Integer.parseInt(c));
                          
            }
        }


        return (int)stack.peek();
    }