[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:

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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:

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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:

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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:

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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:

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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:

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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:

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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?

 

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