https://leetcode.com/problems/add-binary/
I tried to convert the string into integers using String.parseInt(stringname,2) and add the numbers, then i tried to return Integer.toBinaryString(sum).but for large strings i am getting errors.
class Solution {
public String addBinary(String a, String b) {
int n1=Integer.parseInt(a,2);
int n2=Integer.parseInt(b,2);
int n=n1+n2;
return Integer.toBinaryString(n);
}
}
Here is an example.
String addBinary(String a, String b) {
StringBuilder s = new StringBuilder();
char A[] = a.toCharArray(), g;
char B[] = b.toCharArray(), h;
boolean r = false;
int p, q, n = Math.max(p = A.length, q = B.length);
for (int i = 0; i < n; i++) {
g = '0';
h = '0';
if (i < p) g = A[p - 1 - i];
if (i < q) h = B[q - 1 - i];
if (g == '1' && h == '1')
if (r) s.insert(0, '1');
else {
s.insert(0, '0');
r = true;
}
else if (g == '0' && h == '0')
if (!r) s.insert(0, '0');
else {
s.insert(0, '1');
r = false;
}
else s.insert(0, r ? '0' : '1');
if (i + 1 == n && r) s.insert(0, '1');
}
return s.toString();
}
I would just use your approach using BigInteger. I'm using Random to just generate values that are randomly signed for demonstration.
Random r = new Random();
for (int i = 0; i < 10; i++) {
String a = (r.nextBoolean()?"": "-")+Integer.toBinaryString(r.nextInt(100));
String b = (r.nextBoolean()?"": "-")+Integer.toBinaryString(r.nextInt(100));
String result = addBinary(a,b);
System.out.printf("%10s + %10s = %s%n",a,b,result);
}
prints something like
-11000 + -101111 = -1000111
101011 + -101 = 100110
-0 + 110 = 110
-1001101 + -101 = -1010010
-1000 + 10100 = 1100
-111011 + -1001111 = -10001010
110011 + -10100 = 11111
110010 + 1001011 = 1111101
10111 + 1010111 = 1101110
-111001 + 101100 = -1101
The modified method
public static String addBinary(String a, String b) {
BigInteger ba = new BigInteger(a,2);
BigInteger bb = new BigInteger(b,2);
return ba.add(bb).toString(2);
}
Since BigInteger has arbitrary precision, negative binary numbers are shown simply as negative and not in 2's complement form.
