Leetcode--DAY13

Question 1ADD Binary

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

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

Firslty I Wirte A very direct and kind of brute Forch method. Just consider each case of the sum 1, 0, 2, 3 for Anum + Bnum + carry by switch case.

1  Public classSolution {2     3String result = "";4     intCarry = 0;5     6      Publicstring Addbinary (String A, string b) {7         if(a.length () = = 0 )8             returnb;9         Else if(b.length () = = 0)Ten             returnA; One          A          -         intAlen = A.length ()-1; -         intBlen = B.length ()-1; the          -          -          while(Alen >= 0 && Blen >=0){ -             intAnum = A.charat (alen)-' 0 '; +             intBnum = B.charat (blen)-' 0 '; -Calculator (Anum + bnum +carry); +Alen--; ABlen--; at         } -          -          while(Alen >= 0){ -             intAnum = A.charat (alen)-' 0 '; -Calculator (Anum +carry); -Alen--; in         } -          to          while(Blen >= 0){ +             intBnum = B.charat (blen)-' 0 '; -Calculator (Bnum +carry); theBlen--; *         } $         Panax Notoginseng         if(Carry = = 1) -result = "1" +result; the          +         returnresult; A     } the      +      Public voidCalculatorintsum) { -         Switch(sum) { $              Case0: $result = "0" +result; -                  Break; -              Case1: theresult = "1" +result; -Carry = 0;Wuyi                  Break; the              Case2: -result = "0" +result; WuCarry = 1; -                  Break; About              Case3: $result = "1" +result; -Carry = 1; -                  Break; -         } A     } +}

The other-the-does not-use-switch case but use%2 OR/2-defind carry and Sum, which is much clear and shorter.

1  Publicstring Addbinary (String A, string b) {2         if(a.length () = = 0 )3             returnb;4         Else if(b.length () = = 0)5             returnA;6             7String result = "";8         intCarry = 0;9         Ten         intLen =Math.max (A.length (), b.length ()); One          A          for(inti = 0; i < Len; i + +){ -             intAnum = 0; -             intBnum = 0; the             if(I <a.length ()) -Anum = A.charat (A.length ()-1-i)-' 0 '; -             if(I <b.length ()) -Bnum = B.charat (B.length ()-1-i)-' 0 '; +              -             intsum = anum + Bnum +carry; +result = sum%2 +result; Acarry = SUM/2; at         } -          -         if(Carry = = 1) -result = "1" +result; -          -         returnresult; in          -}

Leetcode--DAY13