题解 | #四则运算#

import java.util.Arrays; import java.util.Scanner;

import java.util.Scanner; import java.util.Stack;

public class Main { public static void main(String[] args){ Scanner sc = new Scanner(System.in); String s = sc.nextLine(); int n = s.length(); int num1 = 0; int o1 = 1; int num2 = 1; int o2 = 1; Stack stk = new Stack<>();

    for(int i=0; i<n; i++){
        char c = s.charAt(i);
        if(Character.isDigit(c)){  //遇到数字则定义num2
            int cur = 0;
            while(i<n && Character.isDigit(s.charAt(i))){
                cur = cur * 10 + (s.charAt(i) - '0');
                i++;
            }
            i--;
            num2 = o2 == 1 ? num2 * cur : num2 / cur;
        }else if(c=='*' || c=='/'){  //遇到×÷定义o2
            o2 = c == '*' ? 1 : -1;
        }else if(c == '(' || c=='{' || c=='['){  //遇到括号则保存当前结果，并初始化
            stk.push(num1);
            stk.push(o1);
            stk.push(num2);
            stk.push(o2);

            num1 = 0;
            o1 = 1;
            num2 = 1;
            o2 = 1;
        }else if(c == '+' || c=='-'){  //遇到加减，说明可以开始计算，计算num1并对定义其他几个变量
            if(c=='-' && (i==0 || s.charAt(i-1)=='(' || s.charAt(i-1)=='[' || s.charAt(i-1)=='{')){
                o1 = -1;
                continue;
            }
            num1 = num1 + o1 * num2;
            o1 = (c == '+' ? 1 : -1);
            num2 = 1;
            o2 = 1;
        }else{  //遇到右括号，则出栈，并计算num2
            int cur = num1 + o1 * num2;
            o2 = stk.pop();
            num2 = stk.pop();
            o1 = stk.pop();
            num1 = stk.pop();

            num2 = o2 == 1 ? num2 * cur : num2 / cur;
        }
    }
    System.out.println(num1 + o1 * num2);
}

}