Leetcode -- 227. Basic Calculator II (Stack)

227. Basic Calculator II(Stack)
medium
https://leetcode.com/problems/basic-calculator-ii/

mplement 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

思路： - use stack to store number after operation. initialize sign as '+'
           - calculating current val = val * 10 + char
           - when meet next sign, perform operation of last sign on current number(+,-*,/)
           - finally sum up number in stack
eg: 2+33*2
            (here) sign = +, val = 33, then sign = *

Complexity: Time O(N) Space O(N)

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