Given two binary strings, return their sum (also a binary string).
The input strings are both non-empty and contains only characters 1 or 0.
Example 1:
Input: a = "11", b = "1" Output: "100"
Example 2:
Input: a = "1010", b = "1011" Output: "10101"
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"
1 class Solution {
2 //从最低位加到最高位，当前位相加结果是%2，进位是/2，
3 //记得处理每一次的进位和最后一次的进位，最后反向输出字符。
4 func addBinary(_ a: String, _ b: String) -> String {
5 //存储字符的输出变量
6 var str:String = String()
7 //获取a的字符数量
8 var i:Int = a.count-1
9 //获取b的字符数量
10 var j:Int = b.count-1
11 //进位标志
12 var carry:Int = 0
13 while(i>=0 || j>=0)
14 {
15 var sum:Int = carry
16 if i>=0
17 {
18 //获取该索引的字符
19 var char:Character = a[a.index(a.startIndex, offsetBy: i)]
20 //将字符转换成整数
21 for code in char.unicodeScalars {
22 //字符 0 的ASCII码为48
23 sum += (Int(code.value) - 48)
24 }
25 //自减
26 i-=1
27 }
28 if j>=0
29 {
30 //获取该索引的字符
31 var char:Character = b[b.index(b.startIndex, offsetBy: j)]
32 //将字符转换成整数
33 for code in char.unicodeScalars {
34 //字符 0 的ASCII码为48
35 sum += (Int(code.value) - 48)
36 }
37 //自减
38 j-=1
39 }
40 //转换为临时字符串
41 var temp = String(sum % 2)
42 //追加到字符串末尾
43 str.append(temp[temp.startIndex])
44 carry = sum/2
45 }
46 if carry != 0
47 {
48 //转换为临时字符串
49 var temp = String(carry)
50 //追加到字符串末尾
51 str.append(temp[temp.startIndex])
52 }
53 //返回逆序字符串
54 return String(str.reversed())
55 }
56 }
版本：16ms
1 class Solution {
2 func addBinary(_ a: String, _ b: String) -> String {
3 var reversedResult = ""
4 let aChars = Array(a), bChars = Array(b)
5 var aIndex = aChars.count - 1, bIndex = bChars.count - 1
6 var hasCarryOver = false
7 while aIndex >= 0 || bIndex >= 0 || hasCarryOver {
8 var sum = 0
9 if aIndex >= 0 && aChars[aIndex] == "1" {
10 sum += 1
11 }
12 if bIndex >= 0 && bChars[bIndex] == "1" {
13 sum += 1
14 }
15 if hasCarryOver {
16 sum += 1
17 }
18 reversedResult.append(String(sum % 2))
19 hasCarryOver = sum > 1
20 aIndex -= 1
21 bIndex -= 1
22 }
23 return String(reversedResult.reversed())
24 }
25 }
版本：20ms
1 class Solution {
2 func addBinary(_ a: String, _ b: String) -> String {
3 var str1 = Array(a)
4 var str2 = Array(b)
5
6 var result: [Character] = []
7
8 var hasRemain = false
9 while !(str1.isEmpty) || !(str2.isEmpty) {
10 let first = str1.popLast() ?? "0"
11 let second = str2.popLast() ?? "0"
12
13 switch (first, second) {
14 case ("0", "0"):
15 result.append(hasRemain ? "1" : "0")
16 hasRemain = false
17 case ("0", "1"), ("1", "0"):
18 if hasRemain {
19 result.append("0")
20 } else {
21 result.append("1")
22 }
23 case ("1", "1"):
24 result.append(hasRemain ? "1" : "0")
25 hasRemain = true
26 default: break
27 }
28 }
29
30 if hasRemain { result.append("1") }
31
32 result = result.reversed()
33 return String(result)
34 }
35 }