[LeetCode] Add Binary

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

For example,

a =

"11"

b =

"1"

Return

"100"

.

[Analysis]

二进制字符串加法，先将两个字符串反转，然后进行加法（注意二进制满二进一），最后将结果反转。

[Solution]

class Solution {
public:
string addBinary(string a, string b) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int i, add = 0, sum = 0;
char tmp;
int len1 = a.length();
int len2 = b.length();

// reverse string a
for(i = 0; i < len1/2; ++i){
tmp = a[i];
a[i] = a[len1 - 1 - i];
a[len1 - 1 - i] = tmp;
}

// reverse string b
for(i = 0; i < len2/2; ++i){
tmp = b[i];
b[i] = b[len2 - 1 - i];
b[len2 - 1 - i] = tmp;
}

// add string a and string, store result in res[]
int len = len1 > len2 ? len1 : len2;
char res[len + 2];
for(i = 0; i < len; ++i){
sum = add;
if(i < len1){
sum += (a[i] - '0');
}
if(i < len2){
sum += (b[i] - '0');
}
add = sum / 2;
sum = sum % 2;
res[i] = sum + '0';
}

// be careful the last carry
if(add > 0){
res[i++] = add + '0';
}

// reverse the result
len = i;
for(i = 0; i < len/2; ++i){
tmp = res[i];
res[i] = res[len - 1 - i];
res[len - 1 - i] = tmp;
}

// be careful the '\0'
res[len] = '\0';

return string(res);
}
};

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