二进制求和-LintCode

#ifndef C408_H
#define C408_H
#include<iostream>
#include<string>
#include<cmath>
using namespace std;
//处理字符串,相加进位
class Solution {
public:
/*
* @param a: a number
* @param b: a number
* @return: the result
*/
string addBinary(string &a, string &b) {
// write your code here
int a_size = a.size();
int b_size = b.size();
if (a_size < b_size)
{
int diff = b_size - a_size;
while (diff != 0)
{
a = "0" + a;
diff--;
}
}
else
{
int diff = a_size - b_size;
while (diff != 0)
{
b = "0" + b;
diff--;
}
}
int len = a.size();
int flag = 0;
for (int i = len - 1; i >= 0; --i)
{
if (flag == 0)
{
if (a[i] == '1'&&b[i] == '1')
{
a[i] = '0';
flag = 1;
}
else if (a[i] == '0'&&b[i] == '0')
{
a[i] = '0';
flag = 0;
}
else
{
a[i] = '1';
flag = 0;
}
}
else
{
if (a[i] == '1'&&b[i] == '1')
{
a[i] = '1';
flag = 1;
}
else if (a[i] == '0'&&b[i] == '0')
{
a[i] = '1';
flag = 0;
}
else
{
a[i] = '0';
flag = 1;
}
}
if (i == 0 && flag == 1)
a = "1" + a;
}
return a;
}
};
//将二进制表示的字符串转换为十进制,再进行运算
class Solution2 {
public:
/**
* @param a: a number
* @param b: a number
* @return: the result
*/
string addBinary(string &a, string &b) {
// write your code here
long long size_a = a.size();
long long size_b = b.size();
long long num_a =0, num_b = 0;
for (long long i = 0; i < size_a; ++i)
num_a += (a[i] - '0')*pow(2, size_a - i - 1);
for (long long i = 0; i < size_b; ++i)
num_b += (b[i] - '0')*pow(2, size_b - i - 1);
long long res = num_a + num_b;
string str;
if (res == 0)
return "0";
while (res)
{
if (res & 1 == 1)
str = "1" + str;
else
str = "0" + str;
res = res >> 1;
}
return str;
}
};
#endif