LeetCode-Basic Calculator
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.
Solution:
public class Solution { int index; public int calculate(String s) { if (s.isEmpty()) return 0; index = 0; return calculateRecur("(" + s + ")"); } public int calculateRecur(String s) { if (s.charAt(index) == '(') { index++; } skipSpace(s); int curValue = (s.charAt(index) == '(') ? calculateRecur(s) : getInt(s); while (index < s.length()) { char nextOper = getOperator(s); if (nextOper == ')' || nextOper == 'n') break; int nextNum = -1; if (s.charAt(index) == '(') { nextNum = calculateRecur(s); } else { nextNum = getInt(s); } curValue = operate(curValue, nextNum, nextOper); } skipSpace(s); return curValue; } public int operate(int num1, int num2, char oper) { if (oper == '+') return num1 + num2; if (oper == '-') return num1 - num2; return -1; } public int getInt(String s) { skipSpace(s); int val = 0; while (index < s.length() && s.charAt(index) >= '0' && s.charAt(index) <= '9') { val = val * 10 + (s.charAt(index++) - '0'); } skipSpace(s); return val; } public char getOperator(String s) { skipSpace(s); char oper = 'n'; if (index < s.length()) { oper = s.charAt(index++); } // Clear space for the following possible '('. skipSpace(s); return oper; } public void skipSpace(String s) { while (index < s.length() && s.charAt(index) == ' ') { index++; } } }