# Basic Calculator II

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.

## Solution

1.当前符号比上一个符号优先级高，比如* 高于+，那么直接进栈
2.当前符号低于上一个，那么就要把所有已经在stack里面优先于当前符号的全算完，再推进当前符号
3.当前符号是“（”，直接push
4.当前符号是“）”，就要把所有“（”以前的符号全部算完

## Code

``````public class Solution {
public int calculate(String s) {
if(s==null || s.length()==0) return 0;

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 = cur * 10 + s.charAt(i+1) - '0';
++i;
}
if(!list.isEmpty() && (list.peek() == 2 || list.peek()==3)) {
int op = list.pop();
int opl = list.pop();
int res = 0;
if(op==2) res = opl * cur;
else res = opl / cur;
list.push(res);
} else {
list.push(cur);
}
} else if(c==' ') continue;
else {
switch (c) {
case '+': list.push(0);
break;
case '-': list.push(1);
break;
case '*': list.push(2);
break;
case '/': list.push(3);
break;
default: return -1;
}
}
}

if(list.isEmpty()) return 0;
Collections.reverse(list);

int res = list.poll();

while(!list.isEmpty()) {
int op = list.poll();
int opr = list.poll();
if(op==0) res += opr;
else res -= opr;
}
return res;
}
}
``````