字符串数组中的索引号Swift

Question

I need to check if the index number of a string is divisible by 2.

Here is what I need to do:

I have a string 0BCB7A0D87AD101B500B I need to remove all "0" characters from the string, but only if they are at an odd number in the string. I need to break it down as follows

0B|CB|7A|0D|87|AD|10|1B|50|0B

and only remove the "0" if it is the first character in the pair.

B|CB|7A|D|87|AD|10|1B|50|B

Then put the string back together

BCB7AD87AD101B50B

This code doesn't work, but this is how I am looking at doing it. I'm breaking up the string into an array:

   var characters = Array("0BCB7A0D87AD101B500B")

    for letter in characters {
        if characters(index: Int) % 2 != 0 { // if index has a remainder after being divided by 2
           if letter == "0"{ 
           characters.removeAtIndex(index: Int)
            }
        }

    }

I just can't divide or get the index number of the letter/string in the array.

Answer 1

This would work, it uses a separate variable ( index ) to keep track of the index:

var characters = Array("0BCB7A0D87AD101B500B")
var newCharacters = Array<Character>()
var index = 0
for letter in characters {
    if index % 2 == 1 || letter != "0" {
        newCharacters.append(letter)
    }
    index++
}
var newString = String(newCharacters)

Answer 2

A functional approach to the problem

// zip each character with the index (enumerate)
let charsWithIndex = enumerate("0BCB7A0D87AD101B500B")
// filter away the unwanted characters (filter) and get rid of the indexes (map)
let filteredCharacters = filter(charsWithIndex) { index, char in !(char == "0" && index % 2 == 0) }.map { String($0.1) }
// put the string back together (join)
let filteredString = "".join(filteredCharacters)

Answer 3

Thank ou for your code. I was able to adjust it to remove at the right points.

var characters = Array("0BCB7A0D87AD101B500B")

    var newCharacters = Array<Character>()
    var index = 1
    for letter in characters {
        if index % 2 == 0 {
            newCharacters.append(letter)
        }
        if index % 2 != 0 && letter != "0" {
            newCharacters.append(letter)
        }
        index++
    }
    var newString = String(newCharacters)