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

The input strings are both non-empty and contains only characters 1 or 0 .

Example 1:

Input: a = "11", b = "1"Output: "100"

Example 2:

Input: a = "1010", b = "1011"Output: "10101"

Constraints:

• Each string consists only of '0' or '1' characters.
• 1 <= a.length, b.length <= 10^4
• Each string is either "0" or doesn’t contain any leading zero.

题意：给出两个二进制字符串，返回它们的和（用二进制表示）。

思路

和 LeetCode 2. Add Two Numbers 完全一致：

class Solution {
   public:    string addBinary(string a, string b) {
           string c;        int n = a.size(), m = b.size(), i = n - 1, j = m - 1, carry = 0;        while (i >= 0 || j >= 0) {
               int sum = carry;            if (i >= 0) {
    sum += a[i] - '0'; --i; }            if (j >= 0) {
    sum += b[j] - '0'; --j; };            c.push_back(sum % 2 + '0');            carry = sum / 2;        }        if (carry) c.push_back(carry + '0');        reverse(c.begin(), c.end());        return c;    }};

效率如下：

执行用时：0 ms, 在所有 C++ 提交中击败了100.00% 的用户内存消耗：6.2 MB, 在所有 C++ 提交中击败了98.07% 的用户

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