Leetcode-add Binary

return string  "one""1" "100 ".

This problem is not difficult, pay attention to distinguish the situation.

 Public classSolution { Publicstring Addbinary (String A, string b) {string str=""; intI=a.length ()-1; intJ=b.length ()-1; intPass=0;  while(I >= 0 && j>=0){            if(Character.getnumericvalue (A.charat (i)) +character.getnumericvalue (B.charat (j)) +pass >= 2){                intDigit= (Character.getnumericvalue (A.charat (i)) +character.getnumericvalue (B.charat (j)) +pass)-2; STR=integer.tostring (Digit) +str; Pass=1; I--; J--; }            Else{                intDigit=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){                intDigit=character.getnumericvalue (B.charat (j)) +pass-2; STR=integer.tostring (Digit) +str; Pass=1; J--; }            Else{                intDigit=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){                intDigit=character.getnumericvalue (A.charat (i)) +pass-2; STR=integer.tostring (Digit) +str; Pass=1; I--; }            Else{                intDigit=character.getnumericvalue (A.charat (i)) +Pass; STR=integer.tostring (Digit) +str; Pass=0; I--; }        }        if(pass==1) {str=integer.tostring (pass) +str; }        returnstr; }}

Leetcode-add Binary