Leetcode 224: Basic Calculator, leetcodecalculator
Basic CalculatorTotal Accepted:2801Total Submissions:18103
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-negativeIntegers 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 notUseeval
Built-in library function.
[Idea]
When '(', push the previous results and symbols into the stack. When ')' is encountered, the symbols in the current result * stack are added with the previous results in the stack.
[CODE]
public class Solution { // "1 + 1" public int calculate(String s) { if(s==null || s.length() == 0) return 0; Stack<Integer> stack = new Stack<Integer>(); int res = 0; int sign = 1; for(int i=0; i<s.length(); i++) { char c = s.charAt(i); if(Character.isDigit(c)) { int cur = c-'0'; while(i+1<s.length() && Character.isDigit(s.charAt(i+1))) { cur = 10*cur + s.charAt(++i) - '0'; } res += sign * cur; } else if(c=='-') { sign = -1; } else if(c=='+') { sign = 1; } else if( c=='(') { stack.push(res); res = 0; stack.push(sign); sign = 1; } else if(c==')') { res = stack.pop() * res + stack.pop(); sign = 1; } } return res; }}