[Leetcode 28] 67 Add Binary

Problem:

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

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

Analysis:

Three main steps: 1. add a and b until meet one's end; 2. process the remaining part of the the other string; 3. process the carry bit.

Simulation problem, the time complexity is O(n+m), the space complexity is O(n+m)

Code:

class Solution {
public:
    string addBinary(string a, string b) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int la = a.length()-1, lb = b.length()-1;
        int bitSum, carry = 0, idx=0;
        char *res = new char[100]();
        
        while (la>=0 && lb>=0) {
            bitSum = a[la] + b[lb] + carry - '0' - '0';
            if (bitSum == 2) {//there is a carry
                res[idx++] = '0';
                carry = 1;
            } else if (bitSum == 3) {
                res[idx++] = '1';
                carry = 1;
            } else {
                res[idx++] = bitSum + '0';
                carry = 0;
            }
            
            la--;
            lb--;
        }
        
        while (la>=0) {
            bitSum = a[la] + carry - '0';
            if (bitSum == 2) {//there is a carry
                res[idx++] = '0';
                carry = 1;
            } else {
                res[idx++] = bitSum + '0';
                carry = 0;
            }
            
            la--;
        }
        
        while (lb>=0) {
            bitSum = b[lb] + carry - '0';
            if (bitSum == 2) {//there is a carry
                res[idx++] = '0';
                carry = 1;
            } else {
                res[idx++] = bitSum + '0';
                carry = 0;
            }
            
            lb--;
        }
        
        if (carry > 0) res[idx++] = carry + '0';
        res[idx] = '\0';
        
        for (int i=0, j=idx-1; i <=j; i++, j--) {
            char ch = res[i];
            res[i] = res[j];
            res[j] = ch;
        }

        string sres = res;
        return sres;
    }
};

Attention: