Reputation: 131
Why is '0'
subtracted from n1.charAt(i)
and n2.charAt(j)
?
Question: Given two numbers represented as strings, return multiplication of the numbers as a string.
public String multiply(String num1, String num2) {
String n1 = new StringBuilder(num1).reverse().toString();
String n2 = new StringBuilder(num2).reverse().toString();
int[] d = new int[num1.length()+num2.length()];
//multiply each digit and sum at the corresponding positions
for(int i=0; i<n1.length(); i++){
for(int j=0; j<n2.length(); j++){
d[i+j] += (n1.charAt(i)-'0') * (n2.charAt(j)-'0');
}
}
StringBuilder sb = new StringBuilder();
//calculate each digit
for(int i=0; i<d.length; i++){
int mod = d[i]%10;
int carry = d[i]/10;
if(i+1<d.length){
d[i+1] += carry;
}
sb.insert(0, mod);
}
//remove front 0's
while(sb.charAt(0) == '0' && sb.length()> 1){
sb.deleteCharAt(0);
}
return sb.toString();
}
Upvotes: 2
Views: 816
Reputation: 201537
It's calculating (by subtraction) the numeric value of the character. The character constant 0
is also digit 0
. Consider a loop of the char
value '0' to '9' can also be done as an int
.
for (char ch = '0'; ch <= '9'; ch++) {
System.out.print(ch);
System.out.print(" = " );
System.out.println((int) ch);
}
or
for (int ch = '0'; ch <= '9'; ch++) {
System.out.print((char) ch);
System.out.print(" = " );
System.out.println(ch);
}
which (both) demonstrate the ascii code (technically Unicdoe, but the technique predates Unicode) values for the digits
0 = 48
1 = 49
2 = 50
3 = 51
4 = 52
5 = 53
6 = 54
7 = 55
8 = 56
9 = 57
See also Character.digit(char, int)
which returns the numeric value of the character ch
in the specified radix. Which means you could replace
d[i + j] += (n1.charAt(i) - '0') * (n2.charAt(j) - '0');
with
d[i + j] += Character.digit(n1.charAt(i), 10)
* Character.digit(n2.charAt(j), 10);
of course, the method could be implemented with a call to the Java built-in arbitrary precision type BigInteger
and you could use BigInteger.multiply(BigInteger)
like
return new BigInteger(num1).multiply(new BigInteger(num2)).toString();
Upvotes: 3
Reputation: 7052
When a character is stored in an integer it is converted to it's corresponding ASCII
value stored.
Now the ASCII value of:
'0' (zero as character) is -> 48
'1' (one as character ) is -> 49
......................
and so on...
Now, how can we get the integer out of the character ?
Its easy to subtract 48
from all numeric character and which is ASCII of '0'
Upvotes: 0
Reputation: 1504172
Why is '0' subtracted from n1.charAt(i) and n2.charAt(j) ?
Because the number of the Unicode character representing the digit 0 is 48, not 0.
Imagine you wanted to label 'A' => 0, 'B' => 1 etc... then you'd use
n1.charAt(i) - 'A'
etc... but as it happens, we want '0' => 0, '1' => 1 etc, so we subtract '0'. You need to differentiate between "the value, as an int
" and "the character, as a char
" (where the latter still has a number value, but it's not what you might expect it to be).
Upvotes: 2
Reputation: 394156
Subtracting '0' from a character that contains a digit gives you the numeric value of that character.
For example, '9' - '0' gives you the integer 9.
The reason for that is that the digit characters ('0' to '9') have integer values from 48 to 57, so '9' - '0' is equivalent to 57 - 48, which gives you 9.
Upvotes: 2