题解 | #表达式求值#

和力扣 224. 基本计算器 相同

class Solution {
    private int i = 0;
    public int solve(String s) {
        //数据预处理：去除所有空格
        s = s.replaceAll(" ", "");  
        return calculate(s);
    }

    public int calculate(String s) {
        //s = s.replaceAll(" ", "");    //因为有递归，每次递归都做处理会耗时很久
        //每个递归栈帧中都有自己独享的变量 num, sign, stack
        int num = 0;
        char sign = '+';
        Stack<Integer> stack = new Stack<>();
        for (;i < s.length(); i++) {
            char c = s.charAt(i);
            if (Character.isDigit(c)) num = num * 10 + (c - '0');
            if (c == '(') {
                i++;
                num = calculate(s);    //出现左括号就递归
            }
            if (!Character.isDigit(c) || i == s.length() - 1) {
                switch(sign) {
                    case '+':
                        stack.push(num);
                        break;
                    case '-':
                        stack.push(-num);
                        break;
                    case '*':
                        stack.push(stack.pop() * num);
                        break;
                    case '/':
                        stack.push(stack.pop() / num);
                        break;
                }
                num = 0;
                sign = c;
                if (c == ')') break;    //出现右括号就结束当前 for 循环
            }
        }
        int res = 0;
        while (!stack.isEmpty()) res += stack.pop();    //将当前递归帧中栈内剩余数字相加
        return res;
    }
}