(Stack)Basic Calculator I && II

Basic Calculator I

Implement a basic calculator to evaluate a simple expression string.

The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, non-negative integers and empty spaces .

You may assume that the given expression is always valid.

Some examples:

"1 + 1" = 2
" 2-1 + 2 " = 3
"(1+(4+5+2)-3)+(6+8)" = 23

Note: Do not use the eval built-in library function.

参考leetcode的解法：

// "(1+(4+5+2)-3)+(6+8)" = 23
    public static int calculate(String s) {
        Stack<Integer> stack = new Stack<Integer>();
        int sign = 1;  //表示正负
        int number = 0;  //存储数值
        int result = 0;
        for(int i = 0; i<s.length(); i++){
            char c = s.charAt(i);
            switch (c) {
            //遇到运算符时，把前面的数值带上其前面的运算符加到结果上
            case '+':
                result += sign * number;
                number = 0;
                sign = 1;
                break;
            case '-':
                result += sign * number;
                number = 0;
                sign = -1;
                break;
            case '(':  //c='('时相当于开始一个新表达式，把原来的结果和符号存入栈中
                stack.push(result);
                stack.push(sign);
                result = 0;
                sign = 1;
                number = 0;
                break;
            case ')': //把当前的表达式结果运算完，加上符号累加到原来的结果
                result += sign * number;
                result *= stack.pop();
                result += stack.pop();
                number = 0;
                break;
            case ' ':
                break;
            default:
                number = number * 10 + (c - '0');
                break;
            }
        }
        if(number != 0)
            result += sign * number;
        return result;
    }

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Basic Calculator II

Implement a basic calculator to evaluate a simple expression string.

The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero.

You may assume that the given expression is always valid.

Some examples:

"3+2*2" = 7
" 3/2 " = 1
" 3+5 / 2 " = 5

Note: Do not use the eval built-in library function.

参考leetcode的解法：

// 1+2*3
    public static int calculate(String s) {
        Stack<Integer> stack = new Stack<Integer>();
        char sign = '+';
        int number = ;
        int result = ;
        for(int i=; i<s.length(); i++){
            char c = s.charAt(i);
            if(Character.isDigit(s.charAt(i))){ //读取数值
                number = number *  + c - '';
            }
            //当字符为运算符或者到达表达式末尾时
            if(!Character.isDigit(c) && ' ' != c || i == s.length() - ){
                switch (sign) {  //判断前一个运算符，跟I的本质区别
                case '+':
                    stack.push(number);
                    break;
                case '-':
                    stack.push(-number);
                    break;
                case '*':   //当上一个运算符为*或者/，取出栈顶数进行运算后再放入栈中
                    stack.push(stack.pop()*number);
                    break;
                case '/':
                    stack.push(stack.pop()/number);
                    break;
                }
                number = ;
                sign = c;
            }
        }
        for(Integer i : stack){
            result += i;
        }
        return result;
    }