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

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

思路：

從最低位加到最高位，當前位相加結果是%2，進位是/2，記得處理每一次的進位和最後一次的進位，最後反向輸出字符。

Java:

public class AddBinary {    
    public String addBinary(String a, String b) {
        StringBuilder result = new StringBuilder();
        int pointerA = a.length()-1;
        int pointerB = b.length()-1;
        int carry = 0;
        while(pointerA>=0 || pointerB>=0){
            int sum = carry;
            if(pointerA>=0){
                sum += (a.charAt(pointerA)-‘0‘);
                pointerA--;
            }
            if(pointerB>=0){
                sum += (b.charAt(pointerB)-‘0‘);
                pointerB--;
            }
            result.append(sum%2);
            carry = sum/2;
        }
        if(carry!=0){
            result.append(‘1‘);
        }
        return result.reverse().toString();
    }
}

C++ 1:

class Solution {
public:
    string addBinary(string a, string b) {
        string res;
        size_t res_len = max(a.length(), b.length()) ;

        size_t carry = 0;
        for (int i = 0; i < res_len; ++i) {
            const size_t a_bit_i = i < a.length() ? a[a.length() - 1 - i] - ‘0‘ : 0;
            const size_t b_bit_i = i < b.length() ? b[b.length() - 1 - i] - ‘0‘ : 0;
            size_t sum = carry + a_bit_i + b_bit_i;
            carry = sum / 2;
            sum %= 2;
            res.push_back(‘0‘ + sum);
        }
        if (carry) {
            res.push_back(‘0‘ + carry);
        }
        reverse(res.begin(), res.end());

        return res;
    }
};

C++ 2:

class Solution {
public:
    string addBinary(string a, string b) {
        string res = "";
        int m = a.size() - 1, n = b.size() - 1, carry = 0;
        while (m >= 0 || n >= 0) {
            int p = m >= 0 ? a[m--] - ‘0‘ : 0;
            int q = n >= 0 ? b[n--] - ‘0‘ : 0;
            int sum = p + q + carry;
            res = to_string(sum % 2) + res;
            carry = sum / 2;
        }
        return carry == 1 ? "1" + res : res;
    }
};

Python:

class Solution:
    # @param a, a string
    # @param b, a string
    # @return a string
    def addBinary(self, a, b):
        result, carry, val = "", 0, 0
        for i in xrange(max(len(a), len(b))):
            val = carry
            if i < len(a):
                val += int(a[-(i + 1)])
            if i < len(b): 
                val += int(b[-(i + 1)])
            carry, val = val / 2, val % 2
            result += str(val)
        if carry:
            result += str(carry)
        return result[::-1]

if __name__ == ‘__main__‘:
    result = Solution().addBinary(‘11‘, ‘1‘)
    print result

後續Followup：

如果不是二進制相加，而是十六進制相加呢？只要把算法中的除2和余2換成16，並添加相應的十六進制字母就行了。

如果是帶小數的

[LeetCode] 67. Add Binary 二進制數相加