Question

今天我试着去做我的第一步。我试着编写一个函数，它创建给定列表的所有排列。首先我完全失败了，所以我尝试用python编写函数并逐步翻译它：

蟒：

def get_permutations(elements):
    permutations = []
    if len(elements) == 1:
        return [elements]
    for i in range(len(elements)):
        for perm in get_permutations(elements[0:i] +  elements[i+1:]):
            permutations.append([elements[i]] + perm)
    return permutations

print(get_permutations([1,2,3]))

转到：

func getPermutations(elements []int) [][]int {
    permutations := [][]int{}
    if len(elements) == 1 {
        permutations = [][]int{elements}
        return permutations
    }
    for i := range elements {
        for _, perm := range getPermutations(append(elements[0:i], elements[i+1:]...)) {
            permutations = append(permutations, append([]int{elements[i]}, perm...))
        }
    }
    return permutations
}

func main() {
    x := getPermutations([]int{1, 2, 3})
    fmt.Print(x)
}

虽然python版本创建了这个输出：

[[1,2,3]，[1,3,2]，[2,1,3]，[2,3,1]，[3,1,2]，[3,2,1] ]

go版本创建了这个：

[[3 3 3] [3 3 3] [3 3 3] [3 3 3] [3 3 3] [3 3 3]]

我真的有人可以帮助我。我真的很想知道，我在go代码中做错了什么

Answer 1

getPermutations函数每次迭代都会改变原始的elements切片。在修改之前，您需要复制该切片。

func getPermutations(elements []int) [][]int {
    permutations := [][]int{}
    if len(elements) == 1 {
        permutations = [][]int{elements}
        return permutations
    }
    for i := range elements {
        el := make([]int, len(elements))
        copy(el, elements)

        // or copy via append
        // el := append([]int(nil), elements...)

        for _, perm := range getPermutations(append(el[0:i], el[i+1:]...)) {
            permutations = append(permutations, append([]int{elements[i]}, perm...))
        }
    }
    return permutations
}

https://play.golang.org/p/oewV8iPd8E

去语言创造排列

1 个答案: