Leetcode: Add Binary

题目描述：

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

For example,
a = "11"
b = "1"
Return "100".

 

其实字符串可以直接用下标访问的，我还转成了char* 数组，从两个串的尾部开始相加，记录进位，把结果记录到一个字符数组中。如果最后计算完还有进位，要记得最高位多加一个1。然后将字符数组翻转就是结果了。

 

class Solution {
public:
    string addBinary(string a, string b) {
        if(a.size() == 0)
        {
            return b;
        }
        if(b.size() == 0)
        {
            return a;
        }
        const char* tempa = a.c_str();
        char* aa= new char[strlen(tempa) + 1];
        const char* tempb = b.c_str();
        char* bb= new char[strlen(tempb) + 1];
        strcpy(aa,tempa);
        strcpy(bb,tempb);
        char* cc= new char[strlen(tempa) + strlen(tempb) + 1];
        int lena = strlen(aa);
        int lenb = strlen(bb);
        int i;
        int j = 0;
        int jin = 0;
        if(lena > lenb)
        {
            for(i = lenb - 1;i >= 0; i --)
            {
                int sum = aa[lena - lenb + i] - '0' + bb[i] - '0' + jin;
                if (sum >= 2)
                {
                    jin = 1;
                    cc[j++] = sum - 2 + '0';
                }
                else
                {
                    jin = 0;
                    cc[j++] = sum + '0';
                }
            }
            for(i = lena- lenb - 1;i >= 0;i --)
            {
                int sum = aa[i] - '0' + jin;
                if (sum >= 2)
                {
                    jin = 1;
                    cc[j++] = sum - 2 + '0';
                }
                else
                {
                    jin = 0;
                    cc[j++] = sum + '0';
                }
            }
        }
        else
        {
            for(i = lena - 1;i >= 0; i --)
            {
                int sum = bb[lenb - lena + i] - '0' + aa[i] - '0' + jin;
                if (sum >= 2)
                {
                    jin = 1;
                    cc[j++] = sum - 2 + '0';
                }
                else
                {
                    jin = 0;
                    cc[j++] = sum + '0';
                }
            }
            for(i = lenb- lena - 1;i >= 0;i --)
            {
                int sum = bb[i] - '0' + jin;
                if (sum >= 2)
                {
                    jin = 1;
                    cc[j++] = sum - 2 + '0';
                }
                else
                {
                    jin = 0;
                    cc[j++] = sum + '0';
                }
            }
        }
        if(jin > 0)
        {
            cc[j++] = '1';
        }
        int k = 0;
        char* result = new char[j+1];
        for(i = j-1 ; i >= 0; i --)
        {
            result[k++] = cc[i];
        }
        result[k] = '\0';
        string r = string(result);
        return r;
    }
};