LeetCode

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:

#define __MAIN__
#ifdef __MAIN__
#include <string>
using namespace std;
#endif

class Solution {
public:
    string addBinary(string a, string b) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(a.length() < b.length()){
            return addBinary(b, a);
        }
        
        if(a == "0"){
            return b;
        }else if(b == "0"){
            return a;
        }
        
        int i;
        char *buf = nullptr, tmp;
        int lena, lenb;
        
        lena = a.length();
        lenb = b.length();
        buf = new char[lena + 2];
        for(i = 0; i < lena + 2; ++i){
            buf[i] = 0;
        }
        for(i = 0; i < lena; ++i){
            buf[i] += (a[lena - 1 - i] - '0');
        }
        for(i = 0; i < lenb; ++i){
            buf[i] += (b[lenb - 1 - i] - '0');
        }
        for(i = 0; i < lena + 1; ++i){
            buf[i + 1] += buf[i] / 2;
            buf[i] %= 2;
        }
        for(i = 0; i < lena + 1; ++i){
            buf[i] += '0';
        }
        for(i = 0; i < lena - i; ++i){
            tmp = buf[i];
            buf[i] = buf[lena - i];
            buf[lena - i] = tmp;
        }
        
        string res;
        if(buf[0] > '0'){
            res = string(buf);
        }else{
            res = string(buf + 1);
        }
        delete[] buf;
        
        return res;
    }
};