Alan - 1 year ago 123

Java Question

I have a practice who's task is to find the largest digit in an integer using recursion in java. For example, for the number 13441 the digit '4' will be returned.

I have been trying for a day now and nothing worked.

What I thought could work is the following code, which I can't quite get the "base case" for:

`public static int maxDigit(int n) {`

int max;

if (n/100==0) {

if (n%10>(n/10)%10) {

max=n%10;

}

else

max=(n/10)%10;

}

else if (n%10>n%100)

max=n%10;

else

max=n%100;

return maxDigit(n/10);

}

As you can see it's completely wrong.

Any help would be great. Thank you

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Answer Source

The simplest base case, is that if n is 0, return 0.

```
public static int maxDigit(int n){
if(n==0) // Base case: if n==0, return 0
return 0;
return Math.max(n%10, maxDigit(n/10)); // Return max of current digit and
// maxDigit of the rest
}
```

or, slightly more concise;

```
public static int maxDigit(int n){
return n==0 ? 0 : Math.max(n%10, maxDigit(n/10));
}
```

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