Question： Given two binary strings, return their sum (also a binary string). For example, a = "11" b = "1" Return "100". Anwser 1： [cpp] class Solution { public: string addBinary(string a, string b) { // Start typing your C/C++ solution below // DO NOT write int main() function string sum = ""; int alen = a.length() - 1; int blen = b.length() - 1; int carry = 0; while (alen >= 0 || blen >= 0 || carry > 0) { int v = carry; if (alen >= 0) v += a[alen] - '0'; if (blen >= 0) v += b[blen] - '0'; carry = v / 2; sum = string(1, ( '0' + (v&1) ) ) + sum; alen--; blen--; } return sum; } }; Anwser 2： wrong for large and large binary string, such as a = "111111111111111111111111000000000000000000000000000000000000000000111111111" [cpp] class Solution { public: int str2num(string str){ int len = str.length(); if(len <= 0) return 0; int sum = 0; int base = 2; int pow = 1; for(int i = 1; i <= len; i++){ int val = str[len - i] - '0'; if(i == 1){ sum += val; } else { pow *= base; sum += val * pow; } } return sum; } string num2str(int num){ if(num == 0 ) return "0"; string str = ""; int mod = num % 2; do{ str = string(1, mod + '0') + str; num = num / 2; mod = num % 2; } while (num > 0); return str; } string addBinary(string a, string b) { // Start typing your C/C++ solution below // DO NOT write int main() function return num2str( str2num(a) + str2num(b) ); } }; 注意點： 1） 思路是將binary先轉化成整數（int， long， ulong， long long等），然後相加（a + b），最後再將整數和轉化回binary字符串 2） 對小數據，此方法可行（Judge Small is ok）； 但對大數據，此方法溢出（Judge Large is wrong） 3） 此算法，可以引申為大整數計算問題（大整數加、減、乘、除）
