Given two binary strings, return their sum (also a binary string). For example, a = "11" b = "1" Return "100".
我自己的做法：用StringBuffer, a和b都从末尾扫描到开头，而StringBuffer则是每次存在开头。循环则是只要aindex>=0 or bindext>=0 or carry>0就继续。加法则是分成3部分，依次判断aindex, bindex, carry是否满足加的条件，满足就加。这样写结构非常清晰。可以作为这类题的解题模板
1 public class Solution { 2 public String addBinary(String a, String b) { 3 if (a==null || a.length()==0) return b; 4 if (b==null || b.length()==0) return a; 5 StringBuffer res = new StringBuffer(); 6 int aindex = a.length() - 1; 7 int bindex = b.length() - 1; 8 int digit = 0; 9 int carry = 0; 10 while (aindex>=0 || bindex>=0 || carry>0) { 11 int result = 0; 12 if (aindex >= 0) { 13 result += (int)(a.charAt(aindex) - '0'); 14 aindex--; 15 } 16 if (bindex >= 0) { 17 result += (int)(b.charAt(bindex) - '0'); 18 bindex--; 19 } 20 if (carry > 0) { 21 result += carry; 22 } 23 digit = result % 2; 24 carry = result / 2; 25 res.insert(0, digit); 26 } 27 return res.toString(); 28 } 29 }
别人的代码，用StringBuffer，差不多
1 public String addBinary(String a, String b) { 2 if(a==null || a.length()==0) 3 return b; 4 if(b==null || b.length()==0) 5 return a; 6 int i=a.length()-1; 7 int j=b.length()-1; 8 int carry = 0; 9 StringBuilder res = new StringBuilder(); 10 while(i>=0&&j>=0) 11 { 12 int digit = (int)(a.charAt(i)-'0'+b.charAt(j)-'0')+carry; 13 carry = digit/2; 14 digit %= 2; 15 res.append(digit); 16 i--; 17 j--; 18 } 19 while(i>=0) 20 { 21 int digit = (int)(a.charAt(i)-'0')+carry; 22 carry = digit/2; 23 digit %= 2; 24 res.append(digit); 25 i--; 26 } 27 while(j>=0) 28 { 29 int digit = (int)(b.charAt(j)-'0')+carry; 30 carry = digit/2; 31 digit %= 2; 32 res.append(digit); 33 j--; 34 } 35 if(carry>0) 36 { 37 res.append(carry); 38 } 39 return res.reverse().toString(); 40 }