LeetCode-Add Binary

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

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

这道题不难，注意分清情况。

public class Solution {
    public String addBinary(String a, String b) {
        String str="";
        int i=a.length()-1;
        int j=b.length()-1;
        int pass=0;
        while(i >= 0 && j>=0){
            if(Character.getNumericValue(a.charAt(i))+Character.getNumericValue(b.charAt(j))+pass >= 2){
                int digit=(Character.getNumericValue(a.charAt(i))+Character.getNumericValue(b.charAt(j))+pass)-2;
                str=Integer.toString(digit)+str;
                pass=1;
                i--;
                j--;
            }
            else{
                int digit=Character.getNumericValue(a.charAt(i))+Character.getNumericValue(b.charAt(j))+pass;
                str=Integer.toString(digit)+str;
                pass=0;
                i--;
                j--;
            }
        }
        while(j>=0){
            if(Character.getNumericValue(b.charAt(j))+pass>=2){
                int digit=Character.getNumericValue(b.charAt(j))+pass-2;
                str=Integer.toString(digit)+str;
                pass=1;
                j--;
            }
            else{
                int digit=Character.getNumericValue(b.charAt(j))+pass;
                str=Integer.toString(digit)+str;
                pass=0;
                j--;
            }
        }
        while(i>=0){
                if(Character.getNumericValue(a.charAt(i))+pass>=2){
                int digit=Character.getNumericValue(a.charAt(i))+pass-2;
                str=Integer.toString(digit)+str;
                pass=1;
                i--;
            }
            else{
                int digit=Character.getNumericValue(a.charAt(i))+pass;
                str=Integer.toString(digit)+str;
                pass=0;
                i--;
            }
        }
        if(pass==1){
            str=Integer.toString(pass)+str;
        }
        return str;
        
    }
}