二进制字符创相加,通过进位的方式逐位考虑。也可以把相加的过程抽象成一个函数。
Given two binary strings, return their sum (also a binary string).
For example,
a = "11"
b = "1"
Return "100"
.
方法一:
class Solution {
public:
string addBinary(string a, string b) {
int a_len=a.length();
int b_len=b.length();
if(a_len==0)
return b;
if(b_len==0)
return a;
string result="";
int max=(a_len>b_len?a_len:b_len);
int j=a_len-1;
int k=b_len-1;
int sum=0;
for(int i=0;i<max;i++,j--,k--)
{
if(j>=0)
{
sum+=a[j]-'0';
}
if(k>=0)
sum+=b[k]-'0';
result=(char)((sum&1)+'0')+result;
sum=sum>>1;
}
if(sum>0)
result='1'+result;
return result;
}
};
方法二:
// add a and b and carry, return a + b + carry.
// carry will be updated when add char.
char add_char(char a, char b, char &c)
{
if (a == b && a == '1')
{
char ret = c;
c = '1';
return ret;
}
else if (a == b && a == '0')
{
char ret = c;
c = '0';
return ret;
}
else
{
if (c == '1')
{
return '0';
}
else
{
return '1';
}
}
}
// Given two binary strings, return their sum (also a binary string).
// For example,
// a = "11"
// b = "1"
// Return "100".
string add_binary(string a, string b)
{
// Init result with the longest string.
string result = a.size() > b.size() ? a : b;
// Init carry with '0'.
char carry = '0';
const char *pa = a.data() + a.size() - 1;
const char *pb = b.data() + b.size() - 1;
string::iterator pc = result.begin() + result.size() - 1;
while (pa != a.data() - 1 || pb != b.data() - 1)
{
if (pa == a.data() - 1)
{
*pc-- = add_char('0', *pb--, carry);
}
else if (pb == b.data() - 1)
{
*pc-- = add_char(*pa--, '0', carry);
}
else
{
*pc-- = add_char(*pa--, *pb--, carry);
}
}
if (carry == '1')
{
return "1" + result;
}
return result;
}