#67.Add Binary
Problem statement
Given two binary strings a and b, return their sum as a binary string.
Example 1
Input: a = "11", b = "1"
Output: "100"
Example 2
Input: a = "1010", b = "1011"
Output: "10101"
Explanation
給定兩個二進位制的字串 a 和 b ，將他們相加並返回二進位制的結果
Solution
將兩字串轉換成十進位，接著兩者相加得到總和，總和在轉換成二進位得到答案
public string AddBinary(string a, string b)
{
var x = Convert.ToInt32(a, 2);
var y = Convert.ToInt32(b, 2);
var res = x + y;
return Convert.ToString(res, 2);
}
測試沒問題直接提交，得到與上一題 Plus One 相同的錯誤，果然沒這麼簡單！於是我尋思著上一題的解法，這題用相同邏輯應該不會有問題，我朝著這個方向前進，建立一個 carry 變數來紀錄是否進位，以及 sb 來合併每次執行結果，上一題 Plus One 是對一個數加一，這題是兩個數相加，沒什麼太難的問題，只是有些小細節要注意
public string AddBinary(string a, string b)
{
// 將 a,b 補 0 直到相同長度
int maxLen = a.Length > b.Length ? a.Length : b.Length;
a = a.PadLeft(maxLen, '0');
b = b.PadLeft(maxLen, '0');
// true 表示進位, false 則無
bool carry = false;
StringBuilder sb = new StringBuilder();
int add;
maxLen--;
// 依據各種狀況進行相加
while (maxLen > -1)
{
add = a[maxLen] + b[maxLen];
if (add == 98) // a+b=2
{
if (carry)
sb.Insert(0, '1');
else
{
sb.Insert(0, '0');
carry = true;
}
}
else if (add == 97) // a+b=1
{
if (carry)
sb.Insert(0, '0');
else
sb.Insert(0, '1');
}
else // a+b=0
{
if (carry)
{
sb.Insert(0, '1');
carry = false;
}
else
sb.Insert(0, '0');
}
maxLen--;
}
if (carry) sb.Insert(0, '1');
return sb.ToString();
}
Reference
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