Buttercola: Zenefits: Infix expression

Buttercola: Zenefits: Infix expression

Understand the problem: The problem looks complicated to implement, actually could be memorized in a template way.  If the token is a number, then push it to the num stack.  If the token is a operator, there are 5 cases to consider: 1. If the opStack is empty   -- push into opStack      -- If the top is '(', then push the token into opStack.      -- Else (which means has higher or equal precedence), consume the stack by doing one computation.  3. If the token is * or /      -- If the top is '(', push the token into opStack      -- Else if top has lower precedence ( + or -), push the token into opStack.      -- Else if the top has equal precedence (* or /), consume the stack by doing one computation.  4. If the token is (     -- Push into opStack    -- Doing computations until we see the '(' in the opStack.  Code (Java): public class Solution { public int calculate(String s) { if (s == null || s.length() == 0) { return 0; } // parse the string String delim = "[ ]+"; s = s.

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