Permutations by swapping: Difference between revisions
Added Swift solution
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(Added Swift solution) |
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[2, 1, 4, 3] => +1
[2, 1, 3, 4] => -1
</pre>
=={{header|Swift}}==
<lang swift>// Implementation of Heap's algorithm.
// See https://en.wikipedia.org/wiki/Heap%27s_algorithm#Details_of_the_algorithm
func generate<T>(array: inout [T], output: (_: [T]) -> Void) {
let n = array.count
var c = Array(repeating: 0, count: n)
var i = 1
output(array)
while i < n {
if c[i] < i {
if (i & 1) == 0 {
array.swapAt(0, i)
} else {
array.swapAt(c[i], i)
}
output(array)
c[i] += 1
i = 1
} else {
c[i] = 0
i += 1
}
}
}
func printPermutations<T>(array: inout [T]) {
var even = true
func p(array: [T]) {
print("\(array) \(even ? "+1" : "-1")")
even = !even
}
generate(array: &array, output: p)
}
print("Permutations and signs for three items:")
var a = [0, 1, 2]
printPermutations(array: &a)
print("\nPermutations and signs for four items:")
var b = [0, 1, 2, 3]
printPermutations(array: &b)</lang>
{{out}}
<pre>
Permutations and signs for three items:
[0, 1, 2] +1
[1, 0, 2] -1
[2, 0, 1] +1
[0, 2, 1] -1
[1, 2, 0] +1
[2, 1, 0] -1
Permutations and signs for four items:
[0, 1, 2, 3] +1
[1, 0, 2, 3] -1
[2, 0, 1, 3] +1
[0, 2, 1, 3] -1
[1, 2, 0, 3] +1
[2, 1, 0, 3] -1
[3, 1, 0, 2] +1
[1, 3, 0, 2] -1
[0, 3, 1, 2] +1
[3, 0, 1, 2] -1
[1, 0, 3, 2] +1
[0, 1, 3, 2] -1
[0, 2, 3, 1] +1
[2, 0, 3, 1] -1
[3, 0, 2, 1] +1
[0, 3, 2, 1] -1
[2, 3, 0, 1] +1
[3, 2, 0, 1] -1
[3, 2, 1, 0] +1
[2, 3, 1, 0] -1
[1, 3, 2, 0] +1
[3, 1, 2, 0] -1
[2, 1, 3, 0] +1
[1, 2, 3, 0] -1
</pre>
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