可以说有一个字符串“johngoestoschool”,它应该变成“JoHnGoEsToScHoOl”,并且如果它们之间有特殊字符,则应该忽略它,例如给定字符串“jo $%@#hn ^ goe!st#os&choo)l”的。答案应为“Jo $%@#Hn ^ GoE!sT#oS&cHoO)l”
从this答案,我们为了迭代我们可以做:
let s = "alpha"
for i in s.characters.indices[s.startIndex..<s.endIndex]
{
print(s[i])
}
为什么我们不能在此处打印“i”的值?
当我们执行i.customPlaygroundQuickLook时,它将int 0键入为int4。
所以我的想法是
if (i.customPlaygroundQuickLook == 3) {
s.characters.currentindex = capitalized
}
请帮助
最佳答案
这应该可以解决您的功能,最困难的部分只是检查天气字符是否为字母,使用inout
和replace range可以提供更好的性能:
func altCaptalized(string: String) -> String {
var stringAr = string.characters.map({ String($0) }) // Convert string to characters array and mapped it to become array of single letter strings
var numOfLetters = 0
// Convert string to array of unicode scalar character to compare in CharacterSet
for (i,uni) in string.unicodeScalars.enumerated() {
//Check if the scalar character is in letter character set
if CharacterSet.letters.contains(uni) {
if numOfLetters % 2 == 0 {
stringAr[i] = stringAr[i].uppercased() //Replace lowercased letter with uppercased
}
numOfLetters += 1
}
}
return stringAr.joined() //Combine all the single letter strings in the array into one string
}
关于ios - 如何大写字符串的每个备用字符?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42667153/