LeetCode 67. Add Binary

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

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

// Only need to pay attention that adding is from the end of string. 

This problem is similar to the add one problem. http://blog.csdn.net/github_34333284/article/details/51103310

#include <string>#include <iostream>using namespace std;// a = "11", b = "1" --> "100"string addBinary(string a, string b) {    if(a.size() == 0) return b;    if(b.size() == 0) return a;    int aSize = a.size() - 1;    int bSize = b.size() - 1;    string res = "";    int overflow = 0;    while(aSize >= 0 && bSize >= 0) {        int tmp = a[aSize] - '0' + b[bSize] - '0' + overflow;        overflow = tmp / 2;        res = to_string(tmp % 2) + res;        aSize--;        bSize--;    }    while(aSize >= 0) {        int tmp = a[aSize--] - '0' + overflow;        overflow = tmp / 2;        res = to_string(tmp % 2) + res;    }    while(bSize >= 0) {        int tmp = b[bSize--] - '0' + overflow;        overflow = tmp / 2;        res = to_string(tmp % 2) + res;    }    if(overflow) res = "1" + res;    return res;}int main(void) {    string a = "11";    string b = "1";    string res = addBinary(a, b);    cout << res << endl;}

Maybe It is better to make code clean.

#include <string>#include <iostream>using namespace std;string addBinary(string a, string b) {  if(a.size() == 0) return b;  if(b.size() == 0) return a;  int aEnd = a.size() - 1;  int bEnd = b.size() - 1;  int overflow = 0;  string res = "";  while(aEnd >= 0 || bEnd >= 0) {    int sum = (aEnd < 0 ? 0 : (a[aEnd] - '0')) + (bEnd < 0 ? 0 : (b[bEnd] - '0')) + overflow;    overflow = sum / 2;    sum = sum % 2;    res = to_string(sum) + res;    aEnd--;    bEnd--;  }  if(overflow) res = "1" + res;  return res;}int main(void) {  cout << addBinary("11", "1") << endl;}