leetcode67. 二进制求和

题目：Given two binary strings, return their sum (also a binary string).
For example,
a = "11"
b = "1"
Return "100".

思路：二进制加法和普通加法的思路没什么差别，甚至更简单。所需要注意的细节就是进位。从最低位开始，进位可能伴随计算直到最高位。所以每一位的加法运算都要将上一次的进位加进去。

对于某一位a和b，上一为的进位c，该为求和之后的和是(a+b+c)%2，进位是(a+b+c)/2。

class Solution {
public:
string addBinary(string a, string b) {
if(a.size()==0||a=="0") return b;
if(b.size()==0||b=="0") return a;
int lena=a.size()-1,lenb=b.size()-1,increase=0;
string sum="";
while(lena>=0 && lenb>=0)
{
sum=std::to_string((a[lena]-'0' + b[lenb] - '0' + increase)%2) + sum;
increase=(a[lena]-'0' + b[lenb] - '0' + increase)/2;
lena--;
lenb--;
}
while(lena>=0)
{
sum=std::to_string((increase+a[lena]-'0')%2)+sum;
increase=(a[lena]-'0'+increase)/2;
lena--;
}
while(lenb>=0)
{
sum=std::to_string((increase+b[lenb]-'0')%2)+sum;
increase=(b[lenb]-'0'+increase)/2;
lenb--;
}
if(increase==1)
sum="1"+sum;
return sum;
}
};