LeetCode_ 67 (binary sum)

Real warriors , Dare to face blasting algorithm

Title Description ：
Here are two binary strings , Returns the sum of them （ In binary ）.

Input is Non empty String and contains only numbers 1 and 0.

Example 1:
Input : a = “11”, b = “1”
Output : “100”

Example 2:
Input : a = “1010”, b = “1011”
Output : “10101”

Tips ：
Each string consists only of characters ‘0’ or ‘1’ form .
1 <= a.length, b.length <= 10^4
If the string is not “0” , No leading zeros .

class Solution {
    
    public String addBinary(String a, String b) {
    
        String res="";
        StringBuilder sb=new StringBuilder();
        StringBuilder sbb=new StringBuilder();
        String s;
        if(a.length()<b.length()){
    
            s=a;
            a=b;
            b=s;
        }
        else s=b;
        int len=a.length()-b.length();
        if(len<0)len=-len;
        for(int k=0;k<len;k++){
    
            sbb.append("0");
        }
        s=sbb.toString();
        b=s+b;
        List<Character> list=new ArrayList<>();
        int[] flag=new int[a.length()];
        char cur='0';
        int i=a.length()-1,j=b.length()-1;
        while(i>=0&&j>=0){
    
            if(i==a.length()-1){
                            // Determine the last bit operation 
                if(a.charAt(i)==b.charAt(j)){
               // If equal 
                    list.add('0');                      // Direct addition 0
                    if(a.charAt(i)=='1')flag[i]=1;        // If it is one , Means to carry 
                    else flag[i]=0;                     // Otherwise, it will not carry 
                    i--;j--;
                                              // Enter the next calculation 
                }else if(a.charAt(i)!=b.charAt(j)){
               // If you don't want to wait 
                    list.add('1');                      // For one 
                    flag[i]=0;                          // No carry 
                    i--;j--;

                }
            }
            else{
    
                if(a.charAt(i)!=b.charAt(j)){
               // If it's not equal 
                    cur='1';                            // For one 
                    flag[i]=0;                          // No carry 
                }
                else{
                                       // equal 
                    cur='0';                            // zero 
                    if(a.charAt(i)=='0'){
                     // If two numbers are zero 
                        flag[i]=0;                      // Then it doesn't carry 
                    }else {
                                 // Otherwise, it will be one 
                        flag[i]=1;                      // carry 
                    }
                }
                if(flag[i+1]==1){
                                      // If the previous carry , Then change 
                    if(cur=='0')cur='1';
                    else{
    
                        cur='0';
                        flag[i]=1;
                    }
                }
                list.add(cur);
                i--;
                j--;
            }
        }
        if(flag[0]==1)list.add('1');
        for(int k=list.size()-1;k>=0 ;k--){
    
            sb.append(list.get(k));
        }
        res=sb.toString();
        return res;
    }
}