nyoj 35 表达式求值

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题目：表达式求值

1.png

思路：

Stack<String> ops存操作
Stack<Double> vals存操作数

碰到操作数，如果ops.peek()是否为"*","/",如果是则计算一次;
碰到"+","-"，如果ops.peek()是否为"+","-",如果是则计算一次;
碰到"(","*","/"直接入符号栈;
碰到")"计算到"("为止，如果"("出栈后栈顶为"*","/"则驱除两个，计算一次;

测试数据说明：

"*","/" 应该是从前往后计算，而不是从后往前计算

(1+2)*5.156/54*4.154/47852*41463=((1+2)*5.156/54+4.154)/47852*41463-56*78/32*0.154=

"*","/" 应该是从前往后计算，而不是从后往前计算
```
1-5+3-2=
```
碰到")",计算直到"(",此时当判断"("前是不是为"*","/"若是，则计算一次
```
5*(1-5*8)-1=
```

代码

import java.util.Scanner;import java.util.Stack;public class Main {    public static void main(String[] args) {        Stack<String> ops = new Stack<String>();        Stack<Double> vals = new Stack<Double>();        Scanner sc = new Scanner(System.in);        int T;        T = sc.nextInt();        while (true) {            if (T == 0) break;            T--;            String str = sc.next();            str = str.replace("(", " ( ").replace(")", " ) ").replace("+", " + ").replace("-", " - ").replace("*", " * ").replace("/", " / ").replace("="," = ");            String[] ss = str.split(" ");            for (String s : ss) {                //System.out.println(s);                //处理                if(s.equals(""));                else if(s.equals("=")){                    while(!ops.isEmpty()){                        double val=vals.pop();                        String op=ops.pop();                        if(op.equals("+")) val=val+vals.pop();                        else val=vals.pop()-val;                        vals.push(val);                    }                }                else if (s.equals("(") || s.equals("*")||s.equals("/")) ops.push(s);                else if(s.equals("+")||s.equals("-")){                    if(!ops.isEmpty()&&(ops.peek().equals("+")||ops.peek().equals("-"))){                        double val=vals.pop();                        String op=ops.pop();                        if(op.equals("+")) val=val+vals.pop();                        else if(op.equals("-")) val=vals.pop()-val;                        vals.push(val);                    }                    ops.push(s);                }                else if (s.equals(")")){                    while(!ops.isEmpty()&&!ops.peek().equals("(")){                        double val=vals.pop();                        String op=ops.pop();                        if(op.equals("+")) val=val+vals.pop();                        else if(op.equals("-")) val=vals.pop()-val;                        vals.push(val);                    }                    if(!ops.isEmpty()) ops.pop();                    //作* /的运算                    if(!ops.isEmpty()&&(ops.peek().equals("*")||ops.peek().equals("/"))){                        double val=vals.pop();                        String op=ops.pop();                        if(op.equals("*")) val=val*vals.pop();                        else val=vals.pop()/val;                        vals.push(val);                    }                }else{                    //碰到数字的时候做* / 运算一次，不做+ - 运算                    double val=Double.parseDouble(s);                    if(!ops.isEmpty()&&(ops.peek().equals("*")||ops.peek().equals("/"))){                        String op=ops.pop();                        if(op.equals("*")) val=val*vals.pop();                        else val=vals.pop()/val;                    }                    vals.push(val);                }            }            System.out.printf("%.2f\n",vals.pop());        }    }}

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