Swift permutations of the digits of an Integer











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2
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I want to return all of the permutations of an integer in Swift.



var outputArray = [[Int]]()
func perms ( _ input: [Int], _ output: [Int]) {
if input.count == 0 {
outputArray.append( [ Int( output.map{ String($0) }.joined() )! ] )
}
else {
for i in 0..<input.count {
let current = [input[i]]
let before = Array( input [0..<i] )
let after = Array( input [(i + 1) ..< input.count] )
perms(before + after, current + output)
}
}
}

func permsOfInteger(_ input: Int) -> [[Int]] {
var inp = input
var inpArray = [Int]()
while inp > 0 {
inpArray.append(inp % 10)
inp = inp / 10
}
perms(inpArray, )
return (outputArray)
}


( permsOfInteger(5432) )



I don't want to switch the algorithm (I'm aware of Heap's algorithm) but rather improve my implementation here. Any advice?










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  • 1




    Just a quick clarification - can you elaborate on what it means to permute an integer? Do you mean generate all the permutations of the numerals that make up the number? Like the number 123 would generate "123", "132", "213", "231", "312", "321"?
    – user1118321
    5 hours ago










  • Exactly. That is correct.
    – stevenpcurtis
    5 hours ago















up vote
2
down vote

favorite












I want to return all of the permutations of an integer in Swift.



var outputArray = [[Int]]()
func perms ( _ input: [Int], _ output: [Int]) {
if input.count == 0 {
outputArray.append( [ Int( output.map{ String($0) }.joined() )! ] )
}
else {
for i in 0..<input.count {
let current = [input[i]]
let before = Array( input [0..<i] )
let after = Array( input [(i + 1) ..< input.count] )
perms(before + after, current + output)
}
}
}

func permsOfInteger(_ input: Int) -> [[Int]] {
var inp = input
var inpArray = [Int]()
while inp > 0 {
inpArray.append(inp % 10)
inp = inp / 10
}
perms(inpArray, )
return (outputArray)
}


( permsOfInteger(5432) )



I don't want to switch the algorithm (I'm aware of Heap's algorithm) but rather improve my implementation here. Any advice?










share|improve this question









New contributor




stevenpcurtis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    Just a quick clarification - can you elaborate on what it means to permute an integer? Do you mean generate all the permutations of the numerals that make up the number? Like the number 123 would generate "123", "132", "213", "231", "312", "321"?
    – user1118321
    5 hours ago










  • Exactly. That is correct.
    – stevenpcurtis
    5 hours ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











I want to return all of the permutations of an integer in Swift.



var outputArray = [[Int]]()
func perms ( _ input: [Int], _ output: [Int]) {
if input.count == 0 {
outputArray.append( [ Int( output.map{ String($0) }.joined() )! ] )
}
else {
for i in 0..<input.count {
let current = [input[i]]
let before = Array( input [0..<i] )
let after = Array( input [(i + 1) ..< input.count] )
perms(before + after, current + output)
}
}
}

func permsOfInteger(_ input: Int) -> [[Int]] {
var inp = input
var inpArray = [Int]()
while inp > 0 {
inpArray.append(inp % 10)
inp = inp / 10
}
perms(inpArray, )
return (outputArray)
}


( permsOfInteger(5432) )



I don't want to switch the algorithm (I'm aware of Heap's algorithm) but rather improve my implementation here. Any advice?










share|improve this question









New contributor




stevenpcurtis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I want to return all of the permutations of an integer in Swift.



var outputArray = [[Int]]()
func perms ( _ input: [Int], _ output: [Int]) {
if input.count == 0 {
outputArray.append( [ Int( output.map{ String($0) }.joined() )! ] )
}
else {
for i in 0..<input.count {
let current = [input[i]]
let before = Array( input [0..<i] )
let after = Array( input [(i + 1) ..< input.count] )
perms(before + after, current + output)
}
}
}

func permsOfInteger(_ input: Int) -> [[Int]] {
var inp = input
var inpArray = [Int]()
while inp > 0 {
inpArray.append(inp % 10)
inp = inp / 10
}
perms(inpArray, )
return (outputArray)
}


( permsOfInteger(5432) )



I don't want to switch the algorithm (I'm aware of Heap's algorithm) but rather improve my implementation here. Any advice?







swift






share|improve this question









New contributor




stevenpcurtis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




stevenpcurtis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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edited 14 mins ago









janos

96.8k12124350




96.8k12124350






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asked 7 hours ago









stevenpcurtis

1111




1111




New contributor




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New contributor





stevenpcurtis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






stevenpcurtis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    Just a quick clarification - can you elaborate on what it means to permute an integer? Do you mean generate all the permutations of the numerals that make up the number? Like the number 123 would generate "123", "132", "213", "231", "312", "321"?
    – user1118321
    5 hours ago










  • Exactly. That is correct.
    – stevenpcurtis
    5 hours ago














  • 1




    Just a quick clarification - can you elaborate on what it means to permute an integer? Do you mean generate all the permutations of the numerals that make up the number? Like the number 123 would generate "123", "132", "213", "231", "312", "321"?
    – user1118321
    5 hours ago










  • Exactly. That is correct.
    – stevenpcurtis
    5 hours ago








1




1




Just a quick clarification - can you elaborate on what it means to permute an integer? Do you mean generate all the permutations of the numerals that make up the number? Like the number 123 would generate "123", "132", "213", "231", "312", "321"?
– user1118321
5 hours ago




Just a quick clarification - can you elaborate on what it means to permute an integer? Do you mean generate all the permutations of the numerals that make up the number? Like the number 123 would generate "123", "132", "213", "231", "312", "321"?
– user1118321
5 hours ago












Exactly. That is correct.
– stevenpcurtis
5 hours ago




Exactly. That is correct.
– stevenpcurtis
5 hours ago










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permsOfInteger uses math operations divide and module to convert an Int to an array. By contrast, perms converts digits to strings and joins them to convert an array to Int. I think it's good to be consistent, I would use math operations for converting in both directions. And I would extract these operations to helper methods intToDigits and digitsToInt.





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    up vote
    0
    down vote













    permsOfInteger uses math operations divide and module to convert an Int to an array. By contrast, perms converts digits to strings and joins them to convert an array to Int. I think it's good to be consistent, I would use math operations for converting in both directions. And I would extract these operations to helper methods intToDigits and digitsToInt.





    share

























      up vote
      0
      down vote













      permsOfInteger uses math operations divide and module to convert an Int to an array. By contrast, perms converts digits to strings and joins them to convert an array to Int. I think it's good to be consistent, I would use math operations for converting in both directions. And I would extract these operations to helper methods intToDigits and digitsToInt.





      share























        up vote
        0
        down vote










        up vote
        0
        down vote









        permsOfInteger uses math operations divide and module to convert an Int to an array. By contrast, perms converts digits to strings and joins them to convert an array to Int. I think it's good to be consistent, I would use math operations for converting in both directions. And I would extract these operations to helper methods intToDigits and digitsToInt.





        share












        permsOfInteger uses math operations divide and module to convert an Int to an array. By contrast, perms converts digits to strings and joins them to convert an array to Int. I think it's good to be consistent, I would use math operations for converting in both directions. And I would extract these operations to helper methods intToDigits and digitsToInt.






        share











        share


        share










        answered 5 mins ago









        janos

        96.8k12124350




        96.8k12124350






















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