"Leetcode-Interview algorithm classic-java implementation" "067-add binary (binary addition)"

"067-add binary (binary addition)" "leetcode-Interview algorithm classic-java Implementation" "All topics Directory Index" Original Question

Given binary strings, return their sum (also a binary string).
For example,
A = "11"
b = "1"
Return "100"

Main Topic

Given two binary strings, they are returned and are also binary strings.

Thinking of solving problems

The corresponding two binary strings are converted to the corresponding integer array, from low to high to add, but also to consider the last addition to expand a bit of the case. For details, see Code implementation.

Code Implementation

Algorithm implementation class

 Public classSolution { PublicStringaddbinary(String A, string b) {int[] CA =New int[A.length ()];int[] cb =New int[B.length ()];//Converts the values in the character array to 0 or 1 of the value         for(inti =0; I < a.length (); i++) {Ca[i] = A.charat (i)-' 0 '; }//Converts the values in the character array to 0 or 1 of the value         for(inti =0; I < b.length (); i++) {Cb[i] = B.charat (i)-' 0 '; }//Long length saved with CA        if(Ca.length < Cb.length) {int[] tmp = CA;            CA = CB;        CB = TMP; }intAi = ca.length-1;//character array CA last index subscript        intBI = cb.length-1;//character array CB last index subscript        intcarry =0;//Inferior carry indicator        intResult//Load Results        //calculation such as: 1010101101 + 10100         while(Ai >=0&& Bi >=0{result = Ca[ai] + Cb[bi] + carry; Ca[ai] = result%2; Carry = result/2;            ai--;        bi--; }//Processing the remaining figures         while(Ai >=0) {result = Ca[ai] + carry; Ca[ai] = result%2; Carry = result/2;if(Carry = =0) { Break;        } ai--; }//Converts the value in the character array to 0 or 1 of the characters         for(inti =0; i < ca.length; i++) {Ca[i] + =' 0 '; }//Do not need to expand one        if(Carry = =0) {Char[] ch =New Char[Ca.length]; for(inti =0; i < ca.length; i++) {Ch[i] = (Char) (Ca[i]); }return NewString (CH); }//need to expand one        Else{Char[] ch =New Char[Ca.length +1]; ch[0] =' 1 '; for(inti =0; i < ca.length; i++) {ch[i +1] = (Char) (Ca[i]); }return NewString (CH); }    }}

Evaluation Results

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"Leetcode-Interview algorithm classic-java implementation" "067-add binary (binary addition)"