Swift permutations of the digits of an Integer
up vote
2
down vote
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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?
swift
New contributor
add a comment |
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?
swift
New contributor
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
add a comment |
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?
swift
New contributor
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
swift
New contributor
New contributor
edited 14 mins ago
janos
96.8k12124350
96.8k12124350
New contributor
asked 7 hours ago
stevenpcurtis
1111
1111
New contributor
New contributor
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
add a comment |
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
add a comment |
1 Answer
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oldest
votes
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
.
add a comment |
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oldest
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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
.
add a comment |
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
.
add a comment |
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
.
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
.
answered 5 mins ago
janos
96.8k12124350
96.8k12124350
add a comment |
add a comment |
stevenpcurtis is a new contributor. Be nice, and check out our Code of Conduct.
stevenpcurtis is a new contributor. Be nice, and check out our Code of Conduct.
stevenpcurtis is a new contributor. Be nice, and check out our Code of Conduct.
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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