# Leetcode Reverse Integer problem solution

In this Leetcode Reverse Integer problem solution we have given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-231, 231 - 1], then return 0.

## Problem solution in Python.

```def reverse(x):
sign = -1 if x<0 else 1
res, x = 0, abs(x)
while x:
res = res*10 + (x%10)
x /= 10
# handle the overflow bound
if res > 2**31+1 or res < -2**31-1:
return 0
return res*sign
```

## Problem solution in Java.

```public int reverse(int x) {
boolean neg = false;
if(x<0){
neg = true;
x = -x;
}
long ans = 0;
int maxPow = (int)Math.log10(x);
while(x>0){
ans+= (x%10 * Math.pow(10,maxPow--));
x=x/10;
}
if(ans > Integer.MAX_VALUE){
return 0;
}
if(neg){
return (int)(-ans);
}else{
return (int) ans;
}
}
```

## Problem solution in C++.

```public:
int reverse(int x) {
long res = 0;
while (x != 0) {
res = 10 * res + x % 10;
x /= 10;
}
return (res > INT_MAX || res < INT_MIN) ? 0 : res;
}
};
```

## Problem solution in C.

```int reverse(int x) {

long reval = 0;
for(;x;x/=10)
reval = reval *10 + x%10;

x = reval;
if(reval != x) return 0;

return reval;
}
```