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;}