Given two binary strings, return their sum (also a binary string).
For example,
a = "11"
b = "1"
Return "100".
1 class Solution { 2 public: 3 string addBinary(string a, string b) { 4 string str = ""; 5 int alen = a.size(); 6 int blen = b.size(); 7 if (alen == 0) 8 return b; 9 if (blen == 0) 10 return a; 11 int i = alen-1; 12 int j = blen-1; 13 int flag = 0; 14 while (i>=0&&j>=0) 15 { 16 if ((a[i] - '0' + b[j] - '0') + flag <= 1) 17 { 18 str += ((a[i] - '0' + b[j] - '0') + flag + '0'); 19 flag = 0; 20 i--; 21 j--; 22 } 23 else 24 { 25 if ((a[i] - '0' + b[j] - '0') + flag == 2) 26 { 27 str += "0"; 28 if (i == 0 && j == 0) 29 str += "1"; 30 else 31 flag = 1; 32 } 33 else 34 { 35 str += "1"; 36 if (i == 0 && j == 0) 37 str += "1"; 38 else 39 flag = 1; 40 } 41 i--; j--; 42 } 43 } 44 if (i < 0) 45 { 46 while (j >= 0) 47 { 48 if (b[j] - '0' + flag <= 1) 49 { 50 str += (b[j] - '0' + flag) + '0'; 51 flag=0; 52 } 53 else 54 { 55 str += "0"; 56 if (j == 0) 57 str += "1"; 58 else 59 flag = 1; 60 } 61 j--; 62 } 63 } 64 else 65 { 66 while (i >= 0) 67 { 68 if (a[i] - '0' + flag <= 1) 69 { 70 str += (a[i] - '0' + flag) + '0'; 71 flag=0; 72 } 73 else 74 { 75 str += "0"; 76 if (i == 0) 77 str += "1"; 78 else 79 flag = 1; 80 } 81 i--; 82 } 83 } 84 string str1 = ""; 85 for (int i = str.size()-1; i >= 0; --i) 86 { 87 str1 += str[i]; 88 } 89 return str1; 90 } 91 };
1 class Solution { 2 public: 3 string addBinary(string a, string b) { 4 int ai=a.size()-1,bi=b.size()-1,extra=0,s,r; 5 string ans; 6 while(ai>=0&&bi>=0){ 7 s=a[ai]-'0'+b[bi]-'0'+extra; 8 r=s%2; 9 extra=s/2; 10 ans.push_back(r+'0'); 11 ai--; 12 bi--; 13 } 14 while(ai>=0){ 15 s=a[ai]-'0'+extra; 16 r=s%2; 17 extra=s/2; 18 ans.push_back(r+'0'); 19 ai--; 20 } 21 while(bi>=0){ 22 s=b[bi]-'0'+extra; 23 r=s%2; 24 extra=s/2; 25 ans.push_back(r+'0'); 26 bi--; 27 } 28 if(extra) ans.push_back('1'); 29 reverse(ans.begin(),ans.end()); 30 return ans; 31 } 32 };