【leetcode】Add Binary

Question：

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

For example,
a = "11"
b = "1"
Return "100".

Anwser 1：      

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" 

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字串

2） 對小資料，此方法可行（Judge Small is ok）； 但對大資料，此方法溢出（Judge Large is wrong）

3） 此演算法，可以引申為大整數計算問題（大整數加、減、乘、除）