67. Add Binary
题目
Given two binary strings, return their sum (also a binary string).
For example,
a = "11"
b = "1"
Return "100".
解析
- 自己写的思路比较麻烦，特别注意字符串string的str[0]是从左边开始的，错过几次了
- 主要考虑进位的几种情况
- 有一种是：思路很简单，先把短的字符串补齐，然后逐位相加。
// 67. Add Binary
class Solution_67 {
public:
string addBinary(string a, string b) {
string ret;
int len = (a.size() < b.size()) ? a.size() : b.size();
bool falg = false; //进位标志
reverse(a.begin(), a.end());
reverse(b.begin(), b.end());
for (int i = 0; i < len;i++)
{
if (a[i]=='1'&&b[i]=='1')
{
if (falg)
{
ret.push_back('1');
}
else
{
ret.push_back('0');
}
falg = true;
}
else if ((a[i] == '1'&&b[i] == '0') || (a[i] == '0'&&b[i] == '1'))
{
if (falg)
{
ret.push_back('0');
falg = true;
}
else
{
ret.push_back('1');
falg = false;
}
}
else if (a[i] == '0'&&b[i] == '0')
{
if (falg)
{
ret.push_back('1');
}
else
{
ret.push_back('0');
}
falg = false;
}
}
int len_max = max(a.size(),b.size());
int len_min = min(a.size(),b.size());
for (int i = len_min; i < len_max; i++)
{
if (a.size() >= b.size())
{
if (a[i] == '1'&&falg)
{
ret.push_back('0');
falg = true;
}
else if (a[i] == '0'&&falg)
{
ret.push_back('1');
falg = false;
}
else
{
ret.push_back(a[i]);
}
}
else
{
if (b[i] == '1'&&falg)
{
ret.push_back('0');
falg = true;
}
else if (b[i] == '0'&&falg)
{
ret.push_back('1');
falg = false;
}
else
{
ret.push_back(b[i]);
}
}
}
if (falg)
{
ret.push_back('1');
}
reverse(ret.begin(), ret.end());
return ret;
}
};
题目来源
