要注意的就是我们在处理每个数字结果之前，都已经确定了它前面有的sign。
class Solution {
public int calculate(String s) {
if (s == null || s.length() == 0){
return 0;
}
//"3+2*2"
//stack: 4 3
//num:2
s = s.replaceAll(" ","");//does this line of code has effect on time complexity?
int res = 0;
char sign = '+';
int num = 0;
Stack<Integer> stack = new Stack<>();
for (int i = 0; i < s.length(); i++){
char c = s.charAt(i);
if (Character.isDigit(c)){
num = c - '0';
while (i + 1 < s.length() && Character.isDigit(s.charAt(i+1))){
num = num*10 + s.charAt(i+1) - '0';
i++;
}
}
if (!Character.isDigit(s.charAt(i)) || i == s.length() - 1){
if (sign == '+'){
stack.push(num);
} else if (sign == '-'){
stack.push(-num);
} else if (sign == '*'){
stack.push(stack.pop() * num);
} else if (sign == '/'){
stack.push(stack.pop() / num);
}
sign = s.charAt(i);
num = 0;
}
}
for (int i : stack){
res += i;
}
return res;
}
}
