LeetCode 67. Add Binary

Posted 2020-08-09

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

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

这个题目只要注意各种情况你就成功了一大半，特别要注意的是对进位赋值后可能产生的变化，以及最后一位进位为1时，

我们要把这个1插进来。同时注意字符串的低位是我们数值的高位

class Solution {
public:
    string addBinary(string a, string b) {
        reverse(begin(a), end(a));
        reverse(begin(b), end(b));
        string result;
        char num = ‘0‘;
        int i = 0;
        for (;i < a.size() && i < b.size();++i)
        {
            if (a[i] == ‘0‘&&b[i] == ‘0‘&&num == ‘0‘)
            {
                result.insert(begin(result), ‘0‘);
            }
            if (a[i] == ‘1‘&&b[i] == ‘1‘&&num == ‘1‘)
            {
                result.insert(begin(result), ‘1‘);
            }
            if ((a[i] == ‘0‘&&b[i] == ‘1‘&&num == ‘1‘) || (a[i] == ‘1‘&&b[i] == ‘0‘&&num == ‘1‘) ||
                (a[i] == ‘1‘&&b[i] == ‘1‘&&num == ‘0‘))
            {
                result.insert(begin(result), ‘0‘);
                num = ‘1‘;
            }
            if (a[i] == ‘1‘&&b[i] == ‘0‘&&num == ‘0‘ || a[i] == ‘0‘&&b[i] == ‘1‘&&num == ‘0‘ ||
                a[i] == ‘0‘&&b[i] == ‘0‘&&num == ‘1‘)
            {
                result.insert(begin(result), ‘1‘);
                num = ‘0‘;
            }
        }
        if (i == a.size())
        {
            for (;i < b.size();++i)
            {
                if (b[i] == ‘0‘&&num == ‘0‘)
                    result.insert(begin(result), ‘0‘);
                if (b[i] == ‘0‘&&num == ‘1‘ || b[i] == ‘1‘&&num == ‘0‘)
                {
                    result.insert(begin(result), ‘1‘);
                    num = ‘0‘;
                }
                if (b[i] == ‘1‘&&num == ‘1‘)
                    result.insert(begin(result), ‘0‘);
            }
        }
        else
        {
            for (;i < a.size();++i)
            {
                if (a[i] == ‘0‘&&num == ‘0‘)
                    result.insert(begin(result), ‘0‘);
                if (a[i] == ‘1‘&&num == ‘1‘)
                    result.insert(begin(result), ‘0‘);
                if (a[i] == ‘0‘&&num == ‘1‘ || a[i] == ‘1‘&&num == ‘0‘)
                {
                    result.insert(begin(result), ‘1‘);
                    num = ‘0‘;
                }
                
            }
        }
        if (num == ‘1‘)result.insert(begin(result), ‘1‘);
        return result;
    }
};