我想弄清楚这个递归函数的事情。我有一个有效的非递归演示，但它使用非递归的静态方法。这些函数从“pool_size”中打印出“数字集”的所有组合。如果有人可以，请帮助我使这个函数递归，那会很棒。package mainimport ( "fmt")func combos_of1(pool_size int) { for i := 1; i < pool_size+1; i++ { fmt.Println(i) } fmt.Println("\n")}func combos_of2(pool_size int) { for i := 1; i < pool_size+1; i++ { for j := i + 1; j < pool_size+1; j++ { fmt.Println(i, j) } } fmt.Println("\n")}func combos_of3(pool_size int) { for i := 1; i < pool_size+1; i++ { for j := i + 1; j < pool_size+1; j++ { for k := j + 1; k < pool_size+1; k++ { fmt.Println(i, j, k) } } } fmt.Println("\n")}func main() { combos_of1(10) combos_of2(10) combos_of3(10)}
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慕工程0101907
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例如，
package main
import "fmt"
func rCombinations(p int, n []int, c []int, ccc [][][]int) [][][]int {
if len(n) == 0 || p <= 0 {
return ccc
}
if len(ccc) == 0 {
ccc = make([][][]int, p)
}
p--
for i := range n {
cc := make([]int, len(c)+1)
copy(cc, c)
cc[len(cc)-1] = n[i]
ccc[len(cc)-1] = append(ccc[len(cc)-1], cc)
ccc = rCombinations(p, n[i+1:], cc, ccc)
}
return ccc
}
func Combinations(p int, n []int) [][][]int {
return rCombinations(p, n, nil, nil)
}
func main() {
pools := 3
numbers := []int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
fmt.Println(pools, "pools", "for", "numbers", numbers)
fmt.Println()
nc := 0
c := Combinations(pools, numbers)
fmt.Println("pools:")
d := " digit : "
for i := range c {
fmt.Println(i+1, d)
d = " digits: "
for j := range c[i] {
nc++
fmt.Println(c[i][j], " ")
}
}
fmt.Println()
fmt.Println(nc, "combinations")
}
输出：
3 pools for numbers [1 2 3 4 5 6 7 8 9 10]
pools:
1 digit :
[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
[10]
2 digits:
[1 2]
[1 3]
[1 4]
[1 5]
[1 6]
[1 7]
[1 8]
[1 9]
[1 10]
[2 3]
[2 4]
[2 5]
[2 6]
[2 7]
[2 8]
[2 9]
[2 10]
[3 4]
[3 5]
[3 6]
[3 7]
[3 8]
[3 9]
[3 10]
[4 5]
[4 6]
[4 7]
[4 8]
[4 9]
[4 10]
[5 6]
[5 7]
[5 8]
[5 9]
[5 10]
[6 7]
[6 8]
[6 9]
[6 10]
[7 8]
[7 9]
[7 10]
[8 9]
[8 10]
[9 10]
3 digits:
[1 2 3]
[1 2 4]
[1 2 5]
[1 2 6]
[1 2 7]
[1 2 8]
[1 2 9]
[1 2 10]
[1 3 4]
[1 3 5]
[1 3 6]
[1 3 7]
[1 3 8]
[1 3 9]
[1 3 10]
[1 4 5]
[1 4 6]
[1 4 7]
[1 4 8]
[1 4 9]
[1 4 10]
[1 5 6]
[1 5 7]
[1 5 8]
[1 5 9]
[1 5 10]
[1 6 7]
[1 6 8]
[1 6 9]
[1 6 10]
[1 7 8]
[1 7 9]
[1 7 10]
[1 8 9]
[1 8 10]
[1 9 10]
[2 3 4]
[2 3 5]
[2 3 6]
[2 3 7]
[2 3 8]
[2 3 9]
[2 3 10]
[2 4 5]
[2 4 6]
[2 4 7]
[2 4 8]
[2 4 9]
[2 4 10]
[2 5 6]
[2 5 7]
[2 5 8]
[2 5 9]
[2 5 10]
[2 6 7]
[2 6 8]
[2 6 9]
[2 6 10]
[2 7 8]
[2 7 9]
[2 7 10]
[2 8 9]
[2 8 10]
[2 9 10]
[3 4 5]
[3 4 6]
[3 4 7]
[3 4 8]
[3 4 9]
[3 4 10]
[3 5 6]
[3 5 7]
[3 5 8]
[3 5 9]
[3 5 10]
[3 6 7]
[3 6 8]
[3 6 9]
[3 6 10]
[3 7 8]
[3 7 9]
[3 7 10]
[3 8 9]
[3 8 10]
[3 9 10]
[4 5 6]
[4 5 7]
[4 5 8]
[4 5 9]
[4 5 10]
[4 6 7]
[4 6 8]
[4 6 9]
[4 6 10]
[4 7 8]
[4 7 9]
[4 7 10]
[4 8 9]
[4 8 10]
[4 9 10]
[5 6 7]
[5 6 8]
[5 6 9]
[5 6 10]
[5 7 8]
[5 7 9]
[5 7 10]
[5 8 9]
[5 8 10]
[5 9 10]
[6 7 8]
[6 7 9]
[6 7 10]
[6 8 9]
[6 8 10]
[6 9 10]
[7 8 9]
[7 8 10]
[7 9 10]
[8 9 10]
175 combinations
单个池的变化是：
package main
import "fmt"
func rPool(p int, n []int, c []int, cc [][]int) [][]int {
if len(n) == 0 || p <= 0 {
return cc
}
p--
for i := range n {
r := make([]int, len(c)+1)
copy(r, c)
r[len(r)-1] = n[i]
if p == 0 {
cc = append(cc, r)
}
cc = rPool(p, n[i+1:], r, cc)
}
return cc
}
func Pool(p int, n []int) [][]int {
return rPool(p, n, nil, nil)
}
func main() {
pool := 9
numbers := []int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
p := Pool(pool, numbers)
fmt.Println(pool, "digit pool", "for", "numbers", numbers)
for i := range p {
fmt.Println(p[i])
}
}
输出：
9 digit pool for numbers [1 2 3 4 5 6 7 8 9 10]
[1 2 3 4 5 6 7 8 9]
[1 2 3 4 5 6 7 8 10]
[1 2 3 4 5 6 7 9 10]
[1 2 3 4 5 6 8 9 10]
[1 2 3 4 5 7 8 9 10]
[1 2 3 4 6 7 8 9 10]
[1 2 3 5 6 7 8 9 10]
[1 2 4 5 6 7 8 9 10]
[1 3 4 5 6 7 8 9 10]
[2 3 4 5 6 7 8 9 10]
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