[LeetCode] 67. Add Binary 二进制数相加

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

The input strings are both non-empty and contains only characters 1 or 0.

Example 1:

Input: a = "11", b = "1"
Output: "100"

Example 2:

Input: a = "1010", b = "1011"
Output: "10101"

解法： 从最低位加到最高位，当前位相加结果是%2，进位是/2，记得处理每一次的进位和最后一次的进位，最后反向输出字符。

Java:

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) {
            int sum = carry;
            if (j >= 0) sum += b.charAt(j--) - '0';
            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();
    }
}　　

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();
    }
}

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]

Python:

   class Solution:
        def addBinary(self, a, b):
            if len(a)==0: return b
            if len(b)==0: return a
            if a[-1] == '1' and b[-1] == '1':
                return self.addBinary(self.addBinary(a[0:-1],b[0:-1]),'1')+'0'
            if a[-1] == '0' and b[-1] == '0':
                return self.addBinary(a[0:-1],b[0:-1])+'0'
            else:
                return self.addBinary(a[0:-1],b[0:-1])+'1'　

Python:

class Solution(object):
    def addBinary(self, a, b):
        """
        :type a: str
        :type b: str
        :rtype: str
        """
        res = ''
        carry = 0
        i, j = len(a) - 1, len(b) - 1
        while i >= 0 or j >= 0:
            su = carry
            if i >= 0:
                su += int(a[i])
            if j >= 0:
                su += int(b[j])  
            carry = su / 2
            res += str(su % 2)
            i -= 1
            j -= 1
 
        if carry > 0:
            res += str(carry)
 
        return res[::-1] 　　

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;
    }
};

　　　　

Followup：

如果不是二进制相加，而是十六进制相加呢？只要把算法中的除2和余2换成16，并添加相应的十六进制字母就行了。

如果是带小数的?

 

All LeetCode Questions List 题目汇总