# LeetCode 7. Reverse Integer

### Kaitian Xie ・1 min read

```
public int reverse(int x) {
int rev = 0;
while (x != 0) {
int pop = x % 10;
x /= 10;
if (rev > Integer.MAX_VALUE / 10 || (rev == Integer.MAX_VALUE / 10 && pop > 7)) {
return 0;
}
if (rev < Integer.MIN_VALUE / 10 || (rev == Integer.MIN_VALUE / 10 && pop < -8)) {
return 0;
}
rev = rev * 10 + pop;
}
return rev;
}
```

Reversing an integer is similar to reversing String. We get the last digit `x % 10`

and add it to `rev`

. We repeat this process until `x`

is 0. The two if statements are used to prevent integer overflows.

*Time Complexity*: `O(log(x))`

*Extra Space*: `O(1)`

## A note on integer overflows

A integer occurs when the sum of `rev * 10 + pop`

is greater/less than the maximum value (`Integer.MAX_VALUE`

)/the minimum value(`Integer.MIN_VALUE`

). So we need the check the values of `rev`

and `pop`

.

- If
`rev > Integer.MAX_VALUE/10`

or`rev < Integer.MIN_VALUE/10`

,`rev * 10 + pop`

will overflow. - If
`rev == Integer.MAX_VALUE / 10`

,`pop`

must be less than or equal to 7 because`2^32 - 1 % 10 == 7`

. - If
`rev == Integer.MIN_VALUE / 10`

,`pop`

must be greater than or equal to -8 because`-2^32 % 10 == 8`

.

Classic DEV Post from Oct 3 '18