LeetCode67-二进制求和(C++实现)

1.题目描述

给定两个二进制字符串，返回他们的和（用二进制表示）。输入为非空字符串且只包含数字 1 和 0。

示例 1:
输入: a = "11", b = "1"
输出: "100"

示例 2:
输入: a = "1010", b = "1011"
输出: "10101"

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/add-binary

2.代码实现 

class Solution {
public:
    string addBinary(string a, string b) {
        if(a.length() == 0 && b.length() == 0)
            return "";
        
        if(a.length() == 0)
            return b;
        
        if(b.length() == 0)
            return a;
        
        string res = "";
        int idx_a = a.length() - 1;
        int idx_b = b.length() - 1;
        char flag = '0';
        while(idx_a >= 0 || idx_b >= 0)
        {
            char val;
            if(idx_a >= 0 && idx_b >= 0)
            {
                if(a[idx_a] == '0' && b[idx_b] == '0')
                {
                    val = flag;
                    flag = '0';
                }
                else if((a[idx_a] == '0' && b[idx_b] == '1') || (a[idx_a] == '1' && b[idx_b] == '0'))
                {
                    if(flag == '0')
                        val = '1';
                    else
                        val = '0';
                } 
                else if(a[idx_a] == '1' && b[idx_b] == '1')
                {
                    val = flag;
                    flag = '1';
                }                
            }
            else if(idx_a >= 0 && idx_b < 0)
            {
                if(a[idx_a] == '0')
                {
                    if(flag == '0')
                        val = '0';
                    else
                        val = '1';
                    flag = '0';
                }
                else if(a[idx_a] == '1')
                {
                    if(flag == '0')
                        val = '1';
                    else
                        val = '0';
                }
            }
            else if(idx_b >= 0 && idx_a < 0)
            {
                if(b[idx_b] == '0')
                {
                    if(flag == '0')
                        val = '0';
                    else
                        val = '1';
                    flag = '0';
                }
                else if(b[idx_b] == '1')
                {
                    if(flag == '0')
                        val = '1';
                    else
                        val = '0';
                }
            }
            idx_a--;
            idx_b--;
            res.insert(res.begin(), val);
        }
        if(flag == '1')
           res.insert(res.begin(), flag); 
        return res;
    }
};