LeetCode - Add Binary

Add Binary

2013.12.22 03:31

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

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

Solution:

　　This is a problem about big integer addition, only in that it's binary. My solution is to add them up bit by bit, reverse the char array and return the result as std::string.

　　Time compexity is O(max(m, n)), space complexity is O(max(m, n)), where m and n are the length of two strings.

Accepted code:

 1 #define __MAIN__

 2 #ifdef __MAIN__

 3 #include <string>

 4 using namespace std;

 5 #endif

 6 

 7 class Solution {

 8 public:

 9     string addBinary(string a, string b) {

10         // Note: The Solution object is instantiated only once and is reused by each test case.

11         if(a.length() < b.length()){

12             return addBinary(b, a);

13         }

14         

15         if(a == "0"){

16             return b;

17         }else if(b == "0"){

18             return a;

19         }

20         

21         int i;

22         char *buf = nullptr, tmp;

23         int lena, lenb;

24         

25         lena = a.length();

26         lenb = b.length();

27         buf = new char[lena + 2];

28         for(i = 0; i < lena + 2; ++i){

29             buf[i] = 0;

30         }

31         for(i = 0; i < lena; ++i){

32             buf[i] += (a[lena - 1 - i] - '0');

33         }

34         for(i = 0; i < lenb; ++i){

35             buf[i] += (b[lenb - 1 - i] - '0');

36         }

37         for(i = 0; i < lena + 1; ++i){

38             buf[i + 1] += buf[i] / 2;

39             buf[i] %= 2;

40         }

41         for(i = 0; i < lena + 1; ++i){

42             buf[i] += '0';

43         }

44         for(i = 0; i < lena - i; ++i){

45             tmp = buf[i];

46             buf[i] = buf[lena - i];

47             buf[lena - i] = tmp;

48         }

49         

50         string res;

51         if(buf[0] > '0'){

52             res = string(buf);

53         }else{

54             res = string(buf + 1);

55         }

56         delete[] buf;

57         

58         return res;

59     }

60 };