227. 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.

Credits:

Special thanks to @ts for adding this problem and creating all test cases.
从前向后遍历，用堆栈stack存各个数，根据符号不同对数做相应的调整：
sign == '+' : stack.push(num);

sign == '-' : stack.push(-num);

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

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

最后遍历stack累加。代码如下：

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