# Straightforward 0ms Java Solution

• ``````public int[] plusOne(int[] digits) {
for(int i = digits.length-1; i >= 0; i--){
if(digits[i] == 9){
digits[i] = 0;
if(i == 0){ //hit 0 pos and there is a 9 in it, need to extend array as we hit case of [9,9...]
int[] newNumber = new int[digits.length+1];
System.arraycopy(digits,0,newNumber,1,digits.length);
newNumber[0] = 1;
return newNumber;
}
}else{
digits[i]++;
break;
}
}
return digits;
}
``````

A lot of the solutions keep track of the carry, but there is really no reason to if you just add 1 each time, and break once you've hit a number != 9.

The one special case is where entire array is filled with 9's i.e. [9,9,9,9] and in this case once we hit the first 9 in the array we just copy the array into a new expanded (size+1) array and return.

• arraycopy is not necessary, since all the digits being copied are 0, which array initialization will automatically have done.

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