LeetCode 第 67 题 (Add Binary)

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

For example,
a = “11”
b = “1”
Return “100”.

两个字符串，计算加法。这道题主要是考察对字符串操作的掌握情况。另外，加法要从低位算起，但是输出时要先输出高位。因此，需要将计算结果先存下来，然后再逆序输出。

访问字符串的字符可以采用传统的 c 语言那种数组下标的方式。也可以用 iterator。
下标方式的代码如下：

string addBinary(string a, string b)
{
    int n1 = a.length() - 1;
    int n2 = b.length() - 1;

    char ai, bi, ci, carry = 0;
    stack<char> out;
    while (n1 >= 0 || n2 >= 0)
    {
        ai = (n1 >= 0) ? a[n1--] - ‘0‘ : 0;
        bi = (n2 >= 0) ? b[n2--] - ‘0‘ : 0;

        ci = ai + bi + carry;
        carry = (ci > 1) ? 1 : 0;
        ci = ci & 0x01;
        out.push(ci);
    }
    if(carry > 0) out.push(carry);

    string c;
    while(!out.empty())
    {
        ci = out.top() + ‘0‘;
        c.push_back(ci);
        out.pop();
    }
    return c;
}

iterator 方式有点特殊，对字符串逆序遍历需要用 reverse_iterator。代码如下：

string addBinary(string a, string b)
{
    string::const_reverse_iterator ita = a.crbegin();
    string::const_reverse_iterator itb = b.crbegin();

    char ai, bi, ci, carry = 0;
    stack<char> out;
    while (ita != a.crend() || itb != b.crend())
    {
        ai = (ita != a.crend()) ? *ita++ - ‘0‘ : 0;
        bi = (itb != b.crend()) ? *itb++ - ‘0‘ : 0;

        ci = ai + bi + carry;
        carry = (ci > 1) ? 1 : 0;
        ci = ci & 0x01;
        out.push(ci);
    }
    if(carry > 0) out.push(carry);

    string c;
    while(!out.empty())
    {
        ci = out.top() + ‘0‘;
        c.push_back(ci);
        out.pop();
    }
    return c;
}