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[lintcode] problem 849 – basic calculator iii

admin 11月 13, 2020 0

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 .

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

You may assume that the given expression is always valid. All intermediate results will be in the range of [-2147483648, 2147483647]

Example

"1 + 1" = 2
" 6-4 / 2 " = 4
"2*(5+5*2)/3+(6/2+8)" = 21
"(2+6* 3+5- (3*14/7+2)*5)+3"=-12

Code

public int (String s) {
    int result = 0;
    int curResult = 0;
    int num = 0;
    char lastSign = '+';

    for (int i = 0; i < s.length(); i++) {
        char ch = s.charAt(i);

        if (Character.isDigit(ch)) {
            num = 10 * num + (ch - '0');
        }
        else if (ch == '(') {
            int count = 0;
            int j = i;

            for (; i < s.length(); i++) {
                if (s.charAt(i) == '(')
                    count++;
                else if (s.charAt(i) == ')')
                    count--;

                if (count == 0)
                    break;
            }

            num = calculate(s.substring(j + 1, i));
        }

        if (ch == '+' || ch == '-' || ch == '*' || ch == '/' || i == s.length() - 1) {
            switch (lastSign) {
                case '+': curResult += num; break;
                case '-': curResult -= num; break;
                case '*': curResult *= num; break;
                case '/': curResult /= num; break;
            }

            if (ch == '+' || ch == '-' || i == s.length() - 1) {
                result += curResult;
                curResult = 0;
            }

            lastSign = ch;
            num = 0;
        }
    }

    return result;
}