67. 二进制求和

旧城等待， 2022-09-11 13:18 135阅读 0赞

[力扣![icon-default.png?t=L892][icon-default.png_t_L892]/images/20220828/295a4690af8c468eb59c9c4a6afcdee3.png][icon-default.png_t_L892_icon-default.png_t_L892_https_leetcode-cn.com_problems_add-binary]

难度简单674

给你两个二进制字符串，返回它们的和（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"

**提示：**

*  每个字符串仅由字符 `'0'` 或 `'1'` 组成。
 *  `1 <= a.length, b.length <= 10^4`
 *  字符串如果不是 `"0"` ，就都不含前导零。

通过次数192,102提交次数353,460

class Solution {
        public String addBinary(String a, String b) {
           int lenA = a.length()-1;
           int lenB = b.length()-1;
           String ans = "";
           int temp = 0;
           int flag = 0;
           while(lenA>=0 && lenB>=0)
           {
               if(a.charAt(lenA) == '1') temp++;
               if(b.charAt(lenB) == '1') temp++;
               temp +=flag;
               if(temp==3)
               {
                   temp = 1;
                   flag = 1;
               }
               else if(temp == 2) 
               {
                   temp = 0;
                   flag = 1;
               }
               else if(temp==1)
               {
                   temp = 1;
                   flag = 0;
               } 
               ans = ""+temp+ans;
               temp =0;
               lenA--;
               lenB--;
            //    System.out.println("flag=="+flag);
           }
           while(lenA>=0)
           {
               if(a.charAt(lenA) == '1' && flag ==1)
               {
                   temp = 0;
                   flag = 1;
               } 
               else if(a.charAt(lenA) == '0'&& flag ==0)
               {
                   temp = 0;
                   flag = 0;
               }
               else
               {
                   temp = 1;
                   flag = 0;
               }
               ans = ""+temp+ans;
               temp =0;
               lenA--;
           }
           while(lenB>=0)
           {
               if(b.charAt(lenB) == '1' && flag ==1)
               {
                   temp = 0;
                   flag = 1;
               } 
               else if(b.charAt(lenB) == '0' && flag ==0)
               {
                   temp = 0;
                   flag = 0;
               }
               else
               {
                   temp = 1;
                   flag = 0;
               }
               ans = ""+temp+ans;
               temp =0;
               lenB--;
           }
           if(flag == 1) ans =""+flag+ans;
           return ans;
        }
    }

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[icon-default.png_t_L892]: https://csdnimg.cn/release/blog_editor_html/release1.9.1/ckeditor/plugins/CsdnLink/icons/icon-default.png?t=L892
[icon-default.png_t_L892_icon-default.png_t_L892_https_leetcode-cn.com_problems_add-binary]: /images/20220828/295a4690af8c468eb59c9c4a6afcdee3.png
[watermark_type_ZHJvaWRzYW5zZmFsbGJhY2s_shadow_50_text_Q1NETiBA5Yqq5Yqb5a2m5Lmg55qE5qKF5a2Q_size_18_color_FFFFFF_t_70_g_se_x_16]: /images/20220828/5a62c8f9f4564f8c8064eed73f87339a.png