Generate all permutations in go

问题:

I am looking for a way to generate all possible permutations of a list of elements. Something similar to python's itertools.permutations(arr)

permutations ([])
[]

permutations ([1])
[1]

permutations ([1,2])
[1, 2]
[2, 1]

permutations ([1,2,3])
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]

With the difference that I do not care whether permutations would be generated on demand (like a generator in python) or all together. I also do not care whether they will be lexicographically sorted. All I need is to somehow get these n! permutations.

回答1:

There are a lot of the algorithms that generate permutations. One of the easiest I found is Heap's algorithm:

It generates each permutation from the previous one by choosing a pair of elements to interchange.

The idea and a pseudocode that prints the permutations one after another is outlined in the above link. Here is my implementation of the algorithm which returns all permutations

func permutations(arr []int)[][]int{
    var helper func([]int, int)
    res := [][]int{}

    helper = func(arr []int, n int){
        if n == 1{
            tmp := make([]int, len(arr))
            copy(tmp, arr)
            res = append(res, tmp)
        } else {
            for i := 0; i < n; i++{
                helper(arr, n - 1)
                if n % 2 == 1{
                    tmp := arr[i]
                    arr[i] = arr[n - 1]
                    arr[n - 1] = tmp
                } else {
                    tmp := arr[0]
                    arr[0] = arr[n - 1]
                    arr[n - 1] = tmp
                }
            }
        }
    }
    helper(arr, len(arr))
    return res
}

and here is an example of how to use it (Go playground):

arr := []int{1, 2, 3}
fmt.Println(permutations(arr))
[[1 2 3] [2 1 3] [3 2 1] [2 3 1] [3 1 2] [1 3 2]]

One thing to notice that the permutations are not sorted lexicographically (as you have seen in itertools.permutations). If for some reason you need it to be sorted, one way I have found it is to generate them from a factorial number system (it is described in permutation section and allows to quickly find n-th lexicographical permutation).

P.S. you can also take a look at others people code here and here

回答2:

Here's code that iterates over all permutations without generating them all first. The slice p keeps the intermediate state as offsets in a Fisher-Yates shuffle algorithm. This has the nice property that the zero value for p describes the identity permutation.

package main

import "fmt"

func nextPerm(p []int) {
    for i := len(p) - 1; i >= 0; i-- {
        if i == 0 || p[i] < len(p)-i-1 {
            p[i]++
            return
        }
        p[i] = 0
    }
}

func getPerm(orig, p []int) []int {
    result := append([]int{}, orig...)
    for i, v := range p {
        result[i], result[i+v] = result[i+v], result[i]
    }
    return result
}

func main() {
    orig := []int{11, 22, 33}
    for p := make([]int, len(orig)); p[0] < len(p); nextPerm(p) {
        fmt.Println(getPerm(orig, p))
    }
}

回答3:

don't use recursion! If you want to take advantages of go's concurrency and scalability try something like QuickPerm

package main

import (
    "fmt"
)

func main() {
    for perm := range GeneratePermutations([]int{1,2,3,4,5,6,7}){
        fmt.Println(perm)
    }
}
func GeneratePermutations(data []int) <-chan []int {  
    c := make(chan []int)
    go func(c chan []int) {
        defer close(c) 
        permutate(c, data)
    }(c)
    return c 
}
func permutate(c chan []int, inputs []int){
    output := make([]int, len(inputs))
    copy(output, inputs)
    c <- output

    size := len(inputs)
    p := make([]int, size + 1)
    for i := 0; i < size + 1; i++ {
        p[i] = i;
    }
    for i := 1; i < size;{
        p[i]--
        j := 0
        if i % 2 == 1 {
            j = p[i]
        }
        tmp := inputs[j]
        inputs[j] = inputs[i]
        inputs[i] = tmp
        output := make([]int, len(inputs))
        copy(output, inputs)
        c <- output
        for i = 1; p[i] == 0; i++{
            p[i] = i
        }
    }
}

回答4:

In my case I had a reference to an array, then I've did a few changes in your example:

func generateIntPermutations(array []int, n int, result *[][]int) {
    if n == 1 {
        dst := make([]int, len(array))
        copy(dst, array[:])
        *result = append(*result, dst)
    } else {
        for i := 0; i < n; i++ {
            generateIntPermutations(array, n-1, result)
            if n%2 == 0 {
                // Golang allow us to do multiple assignments
                array[0], array[n-1] = array[n-1], array[0]
            } else {
                array[i], array[n-1] = array[n-1], array[i]
            }
        }
    }
}
numbers := []int{0, 1, 2}
var result [][]int
generateIntPermutations(numbers, len(numbers), &result)

// result -> [[0 1 2] [1 0 2] [2 1 0] [1 2 0] [2 0 1] [0 2 1]]