题目描述

请写一个整数计算器，支持加减乘三种运算和括号。

 

示例2

输入

"(2*(3-4))*5"

返回值

-10

 

 运算符号有优先级,所以使用单调栈可以解决改问题。如下代码，效率比较低，后面优化。

import java.util.Stack;

public class AlgoSolveMe {
    public static void main(String[] args) {
        String s = "(2*(3-4))*5";
        //s="100+100";
        //s = "((10+2)*10-(100-(10+20*10-(2*3)))*10*1*2)-2";  // 2198
        AlgoSolveMe algoSolveMe = new AlgoSolveMe();
        System.out.println(algoSolveMe.solve(s));

    }

    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     * 返回表达式的值
     * @param s string字符串 待计算的表达式
     * @return int整型
     */
    public int solve (String s) {
        // "(2*(3-4))*5"
        Stack<Integer> opValue = new Stack<>();
        Stack<Character> opSign = new Stack<>();

         /*
        ((10+2)*10-(100-(10+20*10-(2*3)))*10*1*2)-2
        (120-(100-(10+20*10-6))*10*1*2)-2
        (120-(100-204)*10*1*2)-2
        (120-(-104)*10*1*2)-2
        (120-(-2080))-2
        2200-2 = 2198
        */

        String temp = "";
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            if(ch == '('){
                opSign.push(ch);
            }else if((ch == '*'|| ch == '/' || ch == '+'|| ch == '-') ) {
                if(opSign.isEmpty() || opSign.peek()=='(') {
                    opSign.push(ch);
                }else if(ch == '+' || ch == '-'){
                    while(!opSign.isEmpty() && opSign.peek() != '(') {
                        doOper(opValue, opSign);
                    }
                    opSign.push(ch);
                }else if((ch == '*' || ch == '/') && (opSign.peek()=='+' || opSign.peek()=='-') ){
                    opSign.push(ch);
                }else if(ch == '*' || ch == '/'){
                    while(!opSign.isEmpty() && opSign.peek() != '(' && opSign.peek()!='+' && opSign.peek()!='-') {
                        doOper(opValue, opSign);
                    }
                    opSign.push(ch);
                }
            }else if(ch == ')') {
                while(opSign.peek() != '(') {
                    doOper(opValue, opSign);
                }
                // 弹出右边括号
                opSign.pop();
            }else{
                temp += ch;
                for (int j = i+1; j < s.length(); j++) {
                    // 判断是否为连续的字符是不是数字
                    if(s.charAt(j)<48 || s.charAt(j)>57) {
                       break;
                    }
                    i++;
                    temp += s.charAt(j);
                }
                opValue.push(Integer.valueOf(temp));
                temp= "";
            }
        }

        while(!opSign.isEmpty()){
            doOper(opValue, opSign);
        }

        return opValue.peek();
    }

    private void doOper(Stack<Integer> opValue,Stack<Character> opSign){
        int op1=0,op2=0;
        op2 = opValue.pop();
        op1 = opValue.pop();

        char ops = opSign.pop();
        int res = 0;
        if (ops == '-') {
            res = op1 - op2;
        } else if (ops == '+') {
            res = op1 + op2;
        } else if (ops == '*') {
            res = op1 * op2;
        } else if (ops == '/') {
            res = op1 / op2;
        }

        opValue.push(res);
    }
}