LeetCode Add Binary

 1 class Solution {

 2 public:

 3     string addBinary(string a, string b) {

 4         string res;

 5         int carry = 0, s = 0;

 6         int apos = a.length() - 1;

 7         int bpos = b.length() - 1;

 8         int astp = 0, bstp = 0;

 9         // skip leading zero(s)

10         for (; astp < apos && a[astp] == '0'; astp++);

11         for (; bstp < bpos && b[bstp] == '0'; bstp++);

12         // a/b pos validation

13         bool apv = apos >= astp, bpv = bpos >= bstp;

14         // addition

15         while(apv || bpv) {

16             if (apv) s += a[apos] - '0';

17             if (bpv) s += b[bpos] - '0';

18             carry = s > 1;

19             s = s & 0x1;

20             res.push_back(s ? '1' : '0');

21             apos--, bpos--, s = carry;

22             apv = apos >= astp, bpv = bpos >= bstp;

23         }

24         if (carry) res.push_back('1');

25         if (res.empty()) {

26             res.push_back('0');

27         } else {

28             reverse(res.begin(), res.end());

29         }

30         return res;

31     }

32 };

没研究过C++里字符串操作的效率问题，直接使用+进行字符串连接并重新赋值回去不知道额外开销大不大，从提交的结果来看和用push_back的没什么区别，有的甚至更快