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There is an arithmetic solution to this problem, without converting to a string. It's probably what the string conversion does internally and will therefore likely be much more efficient.

Quoting Wikipedia:

The digit sum of a number \$x\$ in base \$b\$ is given by $$ \sum_{n=0}^{\lfloor \log_b x\rfloor} \frac{1}{b^n}(x \bmod b^{n + 1} - x \bmod b^n). $$

This StackOverflow answer has an implementation in C# (which is very similar to Java):

sum = 0;
while (n != 0) {
 sum += n % 10;
 n /= 10;
}

In addition, I'd like to comment on the naming of your method. In my opinion, digit doesn't describe its function very well. Instead I'd recommend something similar to digitSum.

Note that this approach doesn't handle negative numbers. They cause the function to go into an infinite loop. To handle that case, you can something like the following:

public static int digitSum(int n) {
 int sum = 0;
 bool isNegative = false;
 if (n < 0) {
 isNegative = true;
 n *= -1;
 }
 
 while (n != 0) {
 sum += n % 10;
 n /= 10;
 }
 
 if (isNegative) {
 return -sum;
 } else {
 return sum;
 }
}

There is an arithmetic solution to this problem, without converting to a string. It's probably what the string conversion does internally and will therefore likely be much more efficient.

Quoting Wikipedia:

The digit sum of a number \$x\$ in base \$b\$ is given by $$ \sum_{n=0}^{\lfloor \log_b x\rfloor} \frac{1}{b^n}(x \bmod b^{n + 1} - x \bmod b^n). $$

This StackOverflow answer has an implementation in C# (which is very similar to Java):

sum = 0;
while (n != 0) {
 sum += n % 10;
 n /= 10;
}

In addition, I'd like to comment on the naming of your method. In my opinion, digit doesn't describe its function very well. Instead I'd recommend something similar to digitSum.

Note that this approach doesn't handle negative numbers. They cause the function to go into an infinite loop. To handle that case, you can something like the following:

public static int digitSum(int n) {
 int sum = 0;
 bool isNegative = false;
 if (n < 0) {
 isNegative = true;
 n *= -1;
 }
 
 while (n != 0) {
 sum += n % 10;
 n /= 10;
 }
 
 if (isNegative) {
 return -sum;
 } else {
 return sum;
 }
}

There is an arithmetic solution to this problem, without converting to a string. It's probably what the string conversion does internally and will therefore likely be much more efficient.

Quoting Wikipedia:

The digit sum of a number \$x\$ in base \$b\$ is given by $$ \sum_{n=0}^{\lfloor \log_b x\rfloor} \frac{1}{b^n}(x \bmod b^{n + 1} - x \bmod b^n). $$

This StackOverflow answer has an implementation in C# (which is very similar to Java):

sum = 0;
while (n != 0) {
 sum += n % 10;
 n /= 10;
}

In addition, I'd like to comment on the naming of your method. In my opinion, digit doesn't describe its function very well. Instead I'd recommend something similar to digitSum.

There is an arithmetic solution to this problem, without converting to a string. It's probably what the string conversion does internally and will therefore likely be much more efficient.

Quoting Wikipedia:

The digit sum of a number \$x\$ in base \$b\$ is given by $$ \sum_{n=0}^{\lfloor \log_b x\rfloor} \frac{1}{b^n}(x \bmod b^{n + 1} - x \bmod b^n). $$

This StackOverflow answer has an implementation in C# (which is very similar to Java):

sum = 0;
while (n != 0) {
 sum += n % 10;
 n /= 10;
}

In addition, I'd like to comment on the naming of your method. In my opinion, digit doesn't describe its function very well. Instead I'd recommend something similar to digitSum.

Note that this approach doesn't handle negative numbers. They cause the function to go into an infinite loop. To handle that case, you can something like the following:

public static int digitSum(int n) {
 int sum = 0;
 bool isNegative = false;
 if (n < 0) {
 isNegative = true;
 n *= -1;
 }
 
 while (n != 0) {
 sum += n % 10;
 n /= 10;
 }
 
 if (isNegative) {
 return -sum;
 } else {
 return sum;
 }
}