[LeetCode#227] Basic Calculator II

Problem:

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.

Analysis:

Unlike the Calculator 1, this quesion is acutally much easier!!! You should do through a very tricky method.
To understand this problem, we should firstly understand the puzzle around this problem.
for equation like "22 - 31 * 52 + 22"
when we reach 31, we do not know whether to use it with 22, thus "22 - 31" or keep it for late usage.
In this situation, we apparenlty should not use it with 22 as "22 - 31", but use it with 52 as "31 * 52".

But we really could not know such information, until we reach "*" and get "52". Can we delay such calculation?
Yes!!!!
Why not decide the usage of 31 until we reach "+" after 52, when we have already know "*" and "52". 
We can do it!!!!

Suppose when reach a sign character "+ - * /", we have already know the number and the sign right before the number. 
(sign) num
---------------------------------------------------------------------------------------------------------
if (sign == '+')
    stack.push(num);

if (sign == '-')
    stack.push(-1 * num);

if (sign == '*')
    stack.push(stack.pop() * num);
if (sign == '/')
    stack.push(stack.pop() / num);

iff the sign is '*' or '/', we combine the preivous two numbers together. " (-31) * 52" as single number. (Just like we do the calculation manually. Right!!!)

----------------------------------------------------------------------------------------------------------
Sklls:
1. how to get the num?
char c = s.charAt(i);
if (Character.isDigit(c)) 
    num = num*10 + (c - '0');
Note: Character.isDigit(c) is very elegant!

2. how to tackle the last num?
if (i == s.length() - 1) {
...
}

3. how to get the number before the current num.
stack.pop()

Solution:

public class Solution {
    public int calculate(String s) {
        if (s == null || s.length() == 0)
            return 0;
        Stack<Integer> stack = new Stack<Integer> ();
        int num = 0;
        char sign = '+';
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (Character.isDigit(c)) 
                num = num*10 + (c - '0');
            if ((c != ' ' && !Character.isDigit(c)) || i == s.length() - 1) {
                if (sign == '+')
                    stack.push(num);
                if (sign == '-')
                    stack.push(-1 * num);
                if (sign == '*')
                    stack.push(stack.pop() * num);
                if (sign == '/')
                    stack.push(stack.pop() / num);
                sign = c;
                num = 0;
            }
        }
        int ret = 0;
        for (int i : stack) {
            ret += i;
        }
        return ret;
    }
}