Question

我有两个字符串需要合并。但是我必须像下面的格式那样合并。

var a = "abc"
var b = "def"

我需要这样做：adbecf。

只是我尝试过的某种情况下的正常a + b。但无法解决。有什么想法吗？

更新：

let stringA = "abc"
 let stringB = "def"


func mergeStrings(a: String, b: String) -> String {

    let val = zip(a,b).reduce("") { (result, arg1) in
    let (char1, char2) = arg1
    return result + "\(char1)\(char2)"
}
return val
}

Answer 1

let a = "abc"
let b = "def"
let c = zip(a,b).reduce("") { (result, arg1) in
    let (char1, char2) = arg1
    return result + "\(char1)\(char2)"
}
print(c)//adbecf

简而言之

let c = zip(a,b).reduce("") { $0 + "\($1.0)\($1.1)" }
print(c)

将zip(::)与两个不等长的字符串一起使用将以忽略其余字符串结尾。

let a = "12345"
let b = "67"
let c = zip(a,b).reduce("") { $0 + "\($1.0)\($1.1)" }
print(c)//"1627"--> 345 is ignored

要解决此问题

if a.count > b.count {
    c = c + String(a[b.endIndex...])
} else if b.count > a.count {
    c = c + String(b[a.endIndex...])
}
print(c)//"1627345"

Answer 2

此代码可以处理长度相等的字符串以及长度不相等的字符串的合并，无论是在a循环内还是更长的情况下（如果b较长时都可以结束

var out = ""
for (n, c) in a.enumerated() {
    out.append(c)
    if n < b.count {
        out.append(b[b.index(b.startIndex, offsetBy: n)])
    }
}
if b.count > a.count {
    out += b.suffix(a.count)
}

忽略任何结尾字符

var out = ""
for (n, c) in a.enumerated() {
    if n < b.count {
        out.append(c)
        out.append(b[b.index(b.startIndex, offsetBy: n)])
    }
}

以其他格式附加字符

2 个答案: