# 227.Basic-Calculator-II

## ้ข็ฎๅฐๅ

https://leetcode.com/problems/basic-calculator-ii/description/

## ้ข็ฎๆ่ฟฐ

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.

Example 1:

Input: "3+2*2"
Output: 7
Example 2:

Input: " 3/2 "
Output: 1
Example 3:

Input: " 3+5 / 2 "
Output: 5

Note:
You may assume that the given expression is always valid.
Do not use the eval built-in library function.

## ไปฃ็ 

### Approach 1:

็จๆ ๆๆฏไธชๆฐๅญๅญ่ตทๆฅ๏ผ้ๅฐ* /ๅผนๅบ่ฎก็ฎๅpush

ๆฏไธชๆฐๅญ็จไธไธชๆไฝ็ฌฆๅๅๅฒ

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

if (!Character.isDigit(s.charAt(i))
&& s.charAt(i) != ' '
|| i == len - 1) { // The last character run twice

if (sign == '+') {
stack.push(num);
} else if (sign == '-') {
stack.push(-num);
} else if (sign == '*') {
int val = stack.pop();
stack.push(val * num);
} else if (sign == '/') {
int val = stack.pop();
stack.push(val / num);
}

sign = s.charAt(i);
num = 0;
}
}

int ans = 0;
for (int val : stack) {
ans += val;
}

return ans;
}
}

Last updated