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();
}
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).
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