LeetCode 67 Add Binary

好久不写博客了，最近有跳槽的打算，所以打算重新刷一遍LeetCode，顺便在这里做一点笔记

https://leetcode.com/problems/add-binary/

#67 的Add Binary,大意就是使用string的方式实现两个二进制字符串的加法，思路很简单，先寻找到较长的那个字符串，之后开始从两个字符串的尾部遍历做加法，之所以需要考虑较长字符串，是因为存在1111 + 10这样的情况

使用vector存放最后的输出结果

#define max(a,b) (((a) > (b)) ? (a) : (b))
#define min(a,b) (((a) < (b)) ? (a) : (b))

class Solution {
public:
    string addBinary(string a, string b) {
        int max_len = max(a.length(), b.length());
        int min_len = min(a.length(), b.length());
        int carry = 0, temp = 0;
        int index;
        vector<char> ret;
        string *long_binary;
        string *shot_binary;
        if (max_len == a.length()){
            long_binary = &a;
            shot_binary = &b;
        }
        else{
            long_binary = &b;
            shot_binary = &a;
        }
        for (int index = 1; index <= max_len; index++){
            if (index > min_len){
                temp = carry + ((*long_binary)[max_len - index] - '0');
            }
            else{
                temp = carry + ((*long_binary)[max_len - index] - '0') + ((*shot_binary)[min_len - index] - '0');
            }
            carry = 0;
            if (temp == 2){
                temp = 0;
                carry = 1;
            }
            else if (temp == 3){
                temp = 1;
                carry = 1;
            }
            ret.insert(ret.begin(), '0' + temp);
        }
        if (carry == 1)
            ret.insert(ret.begin(), '1');
        string result;
        result.insert(result.begin(), ret.begin(), ret.end());
        return result;
    }
};