150. Evaluate Reverse Polish Notation - LeetCode

Question

150. Evaluate Reverse Polish Notation

Solution

2 1 + 3 * 是((2+1)*3)的后缀(postfix)或逆波兰(reverse Polish)记法，计算这个表达式容易想到栈，当见到一个数时就入栈，见到操作符时该运算符作用于从该栈中弹出的两个数上，将所得结果入栈。

public int evalRPN(String[] tokens) {
    Stack<Integer> stack = new Stack<>();
    for (String tmp : tokens) {
        if (tmp.length() > 1) {
            stack.push(Integer.parseInt(tmp));
            continue;
        }
        char c = tmp.charAt(0); // String转char
        int a, b;
        switch (c) {
            case '+':
                b = stack.pop();
                a = stack.pop();
                stack.push(a + b);
                break;
            case '-':
                b = stack.pop();
                a = stack.pop();
                stack.push(a - b);
                break;
            case '*':
                b = stack.pop();
                a = stack.pop();
                stack.push(a * b);
                break;
            case '/':
                b = stack.pop();
                a = stack.pop();
                stack.push(a / b);
                break;
            default:
                stack.push(c - '0'); // char 转 int
        }
    }
    return stack.pop();
}

Reference

• https://gitee.com/okokabcd/leetcode