Say I have and array [4, 1, 8, 5] and another array that corresponds to each object in the first array, say ["Four", "One", "Eight", "Five"]. How can I sort the first array in ascending order while also moving the corresponding object in the second array to the same index in Swift?
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3Why are you maintaining two arrays in the first place if they have mutual dependency? Why not make a dictionary like Key is 4 and value is Four? – NSNoob May 23 '16 at 11:55
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I have one array that consists of dictionaries from JSON. I have another array that consists of ints calculated from values in the JSON dictionaries. I guess I could do the calculations then add a new key to each dictionary and sort that way. – raginggoat May 23 '16 at 11:58
2 Answers
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Doesn't sound like best practice but this will solve your problem:
var numbers = [4,7,8,3]
var numbersString = ["Four","Seven","Eight","Three"]
func bubbleSort<T,Y>(inout numbers:[T],inout _ mirrorArray: [Y], _ comapre : (T,T)->(Bool)) -> () {
let numbersLength = numbers.count
for i in 0 ..< numbersLength {
for j in 1 ..< numbersLength-i {
if comapre(numbers[j-1],numbers[j]) {
swap(&numbers[j-1], &numbers[j])
swap(&mirrorArray[j-1], &mirrorArray[j])
}
}
}
}
bubbleSort(&numbers,&numbersString) { (a, b) -> (Bool) in
a<b
}
print(numbers,numbersString)
*This is generic therefore will work with any type and let you supply the condition
Or Ron
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Using quick sort:
func quicksort_swift(inout a:[Int], inout b:[String], start:Int, end:Int) {
if (end - start < 2){
return
}
let p = a[start + (end - start)/2]
var l = start
var r = end - 1
while (l <= r){
if (a[l] < p){
l += 1
continue
}
if (a[r] > p){
r -= 1
continue
}
let t = a[l]
let t1 = b[l]
a[l] = a[r]
b[l] = b[r]
a[r] = t
b[r] = t1
l += 1
r -= 1
}
quicksort_swift(&a, b: &b, start: start, end: r + 1)
quicksort_swift(&a, b: &b, start: r + 1, end: end)
}
Although, the dictionary solution offered by @NSNoob, should be faster and more elegant.
