67. Add Binary

给定两个二进制字符串，返回它们的和（也是一个二进制字符串）

Given two binary strings, return their sum (also a binary string).

For example,
a = "11"
b = "1"
Return "100".

lx223的解法

public class Solution {
    public String addBinary(String a, String b) {
        StringBuilder sb = new StringBuilder();
        int i = a.length() - 1, j = b.length() -1, carry = 0;
        while (i >= 0 || j >= 0) {//从后往前加，因为后面是低位，前面是高位
            int sum = carry;
            if (j >= 0) sum += b.charAt(j--) - '0';//(j--)返回j，再做--操作
            if (i >= 0) sum += a.charAt(i--) - '0';
            sb.append(sum % 2);
            carry = sum / 2;
        }
        if (carry != 0) sb.append(carry);
        return sb.reverse().toString();
    }
}



我的解法

class Solution {
public String addBinary(String a, String b) {
if (a.isEmpty())
return b;
if (b.isEmpty())
return a;

int carry = 0;
int addA = 0;
int addB = 0;
StringBuilder aB = new StringBuilder(a);
StringBuilder bB = new StringBuilder(b);

aB.reverse();
bB.reverse();

StringBuilder sum = new StringBuilder();

for (int i = 0; i < Math.max(a.length(), b.length()); i++) {
if (i >= a.length())
addA = 0;
else
addA = aB.charAt(i) - '0';

if (i >= b.length())
addB = 0;
else
addB = bB.charAt(i) - '0';

sum.insert(0, (addA + addB + carry) % 2);

if (addA + addB + carry >= 2)
carry = 1;
else
carry = 0;

}
if (carry == 1)
sum.insert(0, 1);
return sum.toString();
}
}