【leetcode刷题笔记】Add Binary

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

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

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题解：简单的二进制加法模拟。a，b的最后以为对齐开始进行加法，用carries保存进位，如果加完后最高位还有进位，那么要在结果的最前面加一个1。

代码如下：

 1 public class Solution {

 2     public String addBinary(String a, String b) {

 3         if(a == null || a.length() == 0)

 4             return b;

 5         if(b == null || b.length() == 0)

 6             return a;

 7         

 8         String s = new String();

 9         int carries = 0;

10         int a_kepeler = a.length()-1;

11         int b_kepeler = b.length()-1;

12         

13         while(a_kepeler >= 0 && b_kepeler >= 0){

14             int sum = carries + a.charAt(a_kepeler) - '0' + b.charAt(b_kepeler) - '0';

15             carries = sum / 2;

16             sum = sum % 2;

17             s = String.valueOf(sum) + s;

18             a_kepeler--;

19             b_kepeler--;

20         }

21         

22         while(a_kepeler >= 0){

23             int sum = carries + a.charAt(a_kepeler) - '0';

24             carries = sum / 2;

25             sum = sum % 2;

26             s = String.valueOf(sum) + s;

27             a_kepeler--;

28         }

29         

30         while(b_kepeler >= 0){

31             int sum = carries + b.charAt(b_kepeler) - '0';

32             carries = sum / 2;

33             sum = sum % 2;

34             s = String.valueOf(sum) + s;

35             b_kepeler--;

36         }

37         

38         if(carries > 0)

39             s = "1" + s;

40         

41         return s;

42     }

43 }

上述代码还可以优化，就是如果a的长度小于b，就把a,b交换，使得a总是较长的那个，那么就可以省略第30~36行的while循环了。