LeetCode Add Binary

Problem

Can be as simple as possible

This problem does not worth a note. I put this one just to remind myself that when you found a solution could be very tedious and cumbersome, probably you should think of a more decent one. This problem is the example.

Ugly but work version 1

    public String addBinary(String a, String b) {
        if (a.length() > b.length()) {
            String s = b;
            b = a;
            a = s;
        }
        if (a.isEmpty()) {
            return b;
        }
        char[] s1 = a.toCharArray();
        char[] s2 = b.toCharArray();
        char[] ret = new char[b.length()+1];
        Arrays.fill(ret, '0');
        int i = s1.length-1;
        int j = s2.length-1;
        int k = ret.length-1;
        for (; i >= 0 || j >= 0 || k >= 0; i--, j--, k--) {
            if (i >= 0) {
                if (s1[i] == '1' && s2[j] == '1' && ret[k] == '1') {
                    ret[k] = '1';
                    ret[k-1] = '1';
                } else if (s1[i] == '0' && s2[j] == '0' && ret[k] == '0') {
                    ret[k] = '0';
                } else if ((s1[i] ^ s2[j] ^ ret[k]) == '0') {
                    ret[k-1] = '1';
                    ret[k] = '0';
                } else {
                    ret[k] = '1';
                }
            } else if (j >= 0) {
                if (s2[j] == ret[k]) {
                    if (s2[j] == '1') {
                        ret[k] = '0';
                        ret[k-1] = '1';
                    } 
                } else {
                    ret[k] = '1';
                }
            }
        }
        if (ret[0] == '0') {
            return new String(ret).substring(1);
        } else {
            return new String(ret);
        }
    }

I totally missed the nature of add operation.

Wow

    public String addBinaryS(String a, String b) {
        char[] s1 = a.toCharArray();
        char[] s2 = b.toCharArray();
        String ans = "";
        int i = s1.length-1;
        int j = s2.length-1;
        int carry = 0;
        while (i >= 0 || j >= 0 || carry == 1) {
            int aVal = (i >= 0) ? s1[i] - '0' : 0;
            int bVal = (j >= 0) ? s2[j] - '0' : 0;
            int sum = aVal + bVal + carry;
            carry = sum / 2;
            sum = sum % 2;
            ans = (sum == 1 ? "1" : "0") + ans;
            i--;
            j--;
        }
        return ans;
    }

This solution perfectly utilize the nature of add operation and therefore simplify the logic.