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Given two binary strings, return their sum (also a binary string).
For example,
a = "11"
b = "1"
Return "100".
Analyse: Using XOR(^) to compute the result of the bit-add and line 22 to reserve the carry bit. Be cautious about the last carry.
1 class Solution { 2 public: 3 string addBinary(string a, string b) { 4 int carry = 0; 5 string result; 6 for(int i = a.length() - 1, j = b.length() - 1; i >= 0 || j >= 0; i--, j--){ 7 int sa, sb; 8 if(i < 0) { 9 sa = 0; 10 sb = b[j] - ‘0‘; 11 } 12 else if(j < 0){ 13 sb = 0; 14 sa = a[i] - ‘0‘; 15 } 16 else{ 17 sa = a[i] - ‘0‘; 18 sb = b[j] - ‘0‘; 19 } 20 char current = sa ^ sb ^ carry + ‘0‘; 21 result = current + result; 22 if((carry + sa + sb) / 2) carry = 1; 23 else carry = 0; 24 } 25 if(carry) result = ‘1‘ + result; 26 return result; 27 } 28 };
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原文地址：http://www.cnblogs.com/amazingzoe/p/4436905.html
