力扣算法题—067二进制求和

给定两个二进制字符串，返回他们的和（用二进制表示）。

输入为非空字符串且只包含数字 1 和 0。

示例 1:

输入: a = "11", b = "1"
输出: "100"

示例 2:

输入: a = "1010", b = "1011"
输出: "10101"

#include "_000库函数.h"

//从末尾一位一位的相加
class Solution {
public:
    string addBinary(string a, string b) {
        if (a.empty())return b;
        if (b.empty())return a;
        int p1 = a.size() - 1,p2=b.size()-1;
        string res = p1 > p2 ? a : b;//选用较长的二进制做基
        int p = p1 > p2 ? p1 : p2;
        int s = 0;//进位
        while (p>=0) {
            if (min(p1, p2) >= 0)//两数相加
                s = a[p1--] - '0' + b[p2--] - '0' + s;
            else
                s = res[p] - '0' + s;//进位想加
            
            res[p--] = s % 2 + '0';
            s = s / 2;
        }
        if (s) {//进位
            a = "1";
            res = a + res;
        }
        return res;
    }
};

//下面这种写法又巧妙又简洁，用了两个指针分别指向a和b的末尾，
//然后每次取出一个字符，转为数字，若无法取出字符则按0处理，
//然后定义进位carry，初始化为0，将三者加起来，对2取余即为当前位的数字，
//对2取商即为当前进位的值，记得最后还要判断下carry，如果为1的话，要
//在结果最前面加上一个1，参见代码如下：
class Solution {
public:
    string addBinary(string a, string b) {
        string res = "";
        int m = a.size() - 1, n = b.size() - 1, carry = 0;
        while (m >= 0 || n >= 0) {
            int p = m >= 0 ? a[m--] - '0' : 0;
            int q = n >= 0 ? b[n--] - '0' : 0;
            int sum = p + q + carry;
            res = to_string(sum % 2) + res;
            carry = sum / 2;
        }
        return carry == 1 ? "1" + res : res;
    }
};
void T067() {
    Solution s;
    string a, b;
    a = "11";
    b = "1";
    cout << s.addBinary(a, b) << endl;
    a = "1010";
    b = "1011";
    cout << s.addBinary(a, b) << endl;
    a = "11010";
    b = "";
    cout << s.addBinary(a, b) << endl;


}