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Given two 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 sum 1, 0, 2, 3 for aNum + bNum + carry by switch case.
1 public class Solution { 2 3 String result = ""; 4 int carry = 0; 5 6 public String addBinary(String a, String b) { 7 if (a.length() == 0 ) 8 return b; 9 else if (b.length() == 0) 10 return a; 11 12 13 int alen = a.length()-1; 14 int blen = b.length()-1; 15 16 17 while (alen >= 0 && blen >=0){ 18 int aNum = a.charAt(alen) - ‘0‘; 19 int bNum = b.charAt(blen) - ‘0‘; 20 calculator(aNum + bNum + carry); 21 alen --; 22 blen --; 23 } 24 25 while (alen >= 0){ 26 int aNum = a.charAt(alen) - ‘0‘; 27 calculator(aNum + carry); 28 alen --; 29 } 30 31 while (blen >= 0){ 32 int bNum = b.charAt(blen) - ‘0‘; 33 calculator(bNum + carry); 34 blen --; 35 } 36 37 if (carry == 1) 38 result = "1" + result; 39 40 return result; 41 } 42 43 public void calculator(int sum){ 44 switch(sum){ 45 case 0: 46 result = "0" + result; 47 break; 48 case 1: 49 result = "1" + result; 50 carry = 0; 51 break; 52 case 2: 53 result = "0" + result; 54 carry = 1; 55 break; 56 case 3: 57 result = "1" + result; 58 carry = 1; 59 break; 60 } 61 } 62 }
The other way does not use switch case but use %2 or /2 to defind carry and sum, which is much clear and shorter.
1 public String addBinary(String a, String b) { 2 if (a.length() == 0 ) 3 return b; 4 else if (b.length() == 0) 5 return a; 6 7 String result = ""; 8 int carry = 0; 9 10 int len = Math.max(a.length(), b.length()); 11 12 for (int i = 0; i < len; i ++){ 13 int aNum = 0; 14 int bNum = 0; 15 if (i < a.length()) 16 aNum = a.charAt(a.length() - 1 - i) - ‘0‘; 17 if (i < b.length()) 18 bNum = b.charAt(b.length() - 1 - i) - ‘0‘; 19 20 int sum = aNum + bNum + carry; 21 result = sum%2 + result; 22 carry = sum / 2; 23 } 24 25 if (carry == 1) 26 result = "1" + result; 27 28 return result; 29 30 }
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原文地址：http://www.cnblogs.com/timoBlog/p/4662843.html
