Leetcode9: Add Binary

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

For example,
a = "11"
b = "1"

Return "100".

看看自己写的代码，渣的不行。把string先转成数字，相加后再对每一位判断是否大于1，最后生成的数字再转成string，真是完全不把string当回事啊~~为什么不直接在string上操作呢？而且转化为int，还有上限的问题，比如输入这两个"1110110101", "1110111011" ，结果是："-1226763192"而不是"11101110000"！！真是被自己蠢哭~~

class Solution {public:    string addBinary(string a, string b) {        int Inta = atoi(a.c_str());        int Intb = atoi(b.c_str());        int c = Inta + Intb;                vector<int> temp;        int flag = 0;        int d;        while(c)        {            d = c % 10;            flag = d / 2;            temp.push_back(d%2);            c /= 10;            c += flag;        }                int size = temp.size();        d = 0;        for(int i = size-1; i >= 0; i--)        {            d = d*10 + temp[i];        }                char str[size];        sprintf(str, "%d", d);        //_itoa(d, str, 10);        string tostr(str);        return tostr;    }};

看看别人写的代码：

class Solution {public:    string addBinary(string a, string b) {     string longStr = a.length() > b.length() ? a : b;     string shortStr = a.length() > b.length() ? b : a;      int ad = 0;     int j = longStr.length() - 1;      for(int i = shortStr.length() - 1; i > -1; i--, j--) {         if(shortStr[i] - '0' + longStr[j] - '0' + ad > 1) {             longStr[j] = '0' + (shortStr[i] - '0' + longStr[j] - '0' + ad) % 2;             ad = 1;         }         else {             longStr[j] = '0' + shortStr[i] - '0' + longStr[j] - '0' + ad;             ad = 0;         }     }      for(; j > -1; j--) {         if(longStr[j] - '0' + ad > 1) {             longStr[j] = '0' + (longStr[j] -'0' + ad) % 2;             ad = 1;         }         else {             longStr[j] = '0' + (longStr[j] - '0' + ad) % 2;             ad = 0;         }     }      if(ad) {         longStr.insert(0,1,'1');     }      return longStr;    }};

自己对string的操作太不熟悉了

差距~~~

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