Reputation: 8995
iOS 11, Swift 4.0
Trying to write a recursive function to show all the possible combinations of a string. I got this, but its not quite right since I get only 20 pairs and I should get 24. I cannot see what I have missed here.
Where this coding going wrong?
var ans:Set<String>!
override func viewDidLoad() {
super.viewDidLoad()
// Do any additional setup after loading the view, typically from a nib.
let str = "ABCD"
ans = []
recursiveString(s2s: str, char2s: 0)
print("\(ans) \(ans.count)")
}
func recursiveSwap(s2x: String, c2x: Int, j2m: Int) {
var anschr = Array(s2x)
let tmpchr = anschr[c2x]
anschr[c2x] = anschr[c2x+j2m]
anschr[c2x+j2m] = tmpchr
print("\(String(anschr))")
ans.insert(String(anschr))
if (c2x + j2m + 1) < s2x.count {
recursiveSwap(s2x: String(s2x), c2x: c2x, j2m: j2m+1)
} else {
if (c2x + 1) < s2x.count - 1 {
recursiveSwap(s2x: String(anschr), c2x: c2x + 1, j2m: 1)
}
}
}
func recursiveString(s2s: String, char2s: Int) {
let blue = shiftString(s2s: s2s)
if char2s < s2s.count {
recursiveSwap(s2x: blue, c2x: 0, j2m: 1)
recursiveString(s2s: blue, char2s: char2s + 1)
}
}
func shiftString(s2s: String) -> String {
let str2s = Array(s2s)
let newS = str2s.suffix(str2s.count - 1) + str2s.prefix(1)
return String(newS)
}
It give me ...
CBDA DCBA ACDB ADCB ABDC ABCD DCAB ADCB BDAC BADC BCAD BCDA ADBC BADC CABD CBAD CDBA CDAB BACD CBAD DBCA DCBA DACB DABC
Upvotes: 4
Views: 1358
Reputation: 236260
Not a direct answer to your question but you can get all permutations (translated from java to Swift) as follow:
public extension RangeReplaceableCollection {
func permutations() -> [SubSequence] {
isEmpty ? [] : permutate(.init())
}
private func permutate(_ subSequence: SubSequence) -> [SubSequence] {
var permutations = isEmpty ? [subSequence] : []
indices.forEach {
permutations += (self[..<$0] + self[$0...].dropFirst())
.permutate(subSequence + CollectionOfOne(self[$0]))
}
return permutations
}
}
let str = "ABCD"
print(str.permutations()) // "["ABCD", "ABDC", "ACBD", "ACDB", "ADBC", "ADCB", "BACD", "BADC", "BCAD", "BCDA", "BDAC", "BDCA", "CABD", "CADB", "CBAD", "CBDA", "CDAB", "CDBA", "DABC", "DACB", "DBAC", "DBCA", "DCAB", "DCBA"]\n"
Per-mutating a substring
print("ABCD".dropLast().permutations()) // ["ABC", "ACB", "BAC", "BCA", "CAB", "CBA"]\n"
Upvotes: 4
Reputation: 5160
Since you said: Trying to write a recursive function to show all the possible combinations of a string
I think you could so something like this:
// Takes any collection of T and returns an array of permutations
func permute<C: Collection>(items: C) -> [[C.Iterator.Element]] {
var scratch = Array(items) // This is a scratch space for Heap's algorithm
var result: [[C.Iterator.Element]] = [] // This will accumulate our result
// Heap's algorithm
func heap(_ n: Int) {
if n == 1 {
result.append(scratch)
return
}
for i in 0..<n-1 {
heap(n-1)
let j = (n%2 == 1) ? 0 : i
scratch.swapAt(j, n-1)
}
heap(n-1)
}
// Let's get started
heap(scratch.count)
// And return the result we built up
return result
}
// We could make an overload for permute() that handles strings if we wanted
// But it's often good to be very explicit with strings, and make it clear
// that we're permuting Characters rather than something else.
let string = "ABCD"
let perms = permute(string.characters) // Get the character permutations
let permStrings = perms.map() { String($0) } // Turn them back into strings
print(permStrings) // output if you like
Extracted from this answer which helped me a lot once: Calculate all permutations of a string in Swift
This would give you a number of permutations.
I hope this helps you.
Upvotes: 1