- 用途描述：一组数组
[1, 2, 3]，需计算出该数组所有不交叉顺序的组合
[1]
[2]
[3]
[1,2]
[2,3]
[1,3]
[1,2,3]
- 穷举算法
func getArrayExhaustivity(arr []string) (result [][]string) {
var l = int(math.Pow(float64(2), float64(len(arr))) - 1)
var t []string
for i := 1; i <= l; i++ {
s := i
t = []string{}
for k := 0; s > 0; k++ {
if s&1 == 1 {
t = append(t, arr[k])
}
s >>= 1
}
result = append(result, t)
}
return
}
- 示例代码1
func TestArrayExhaustivity(t *testing.T) {
result := getArrayExhaustivity([]string{"1", "2", "3"})
str, err := json.Marshal(result)
if err != nil {
t.Fatal(err)
}
println(string(str))
}
- 示例结果1
=== RUN TestArrayExhaustivity
[["1"],["2"],["1","2"],["3"],["1","3"],["2","3"],["1","2","3"]]
--- PASS: TestArrayExhaustivity (0.00s)
PASS
Process finished with exit code 0
- 示例代码2
func TestArrayExhaustivity(t *testing.T) {
result := getArrayExhaustivity([]string{"1", "2", "3", "4"})
str, err := json.Marshal(result)
if err != nil {
t.Fatal(err)
}
println(string(str))
}
- 示例结果2
=== RUN TestArrayExhaustivity
[["1"],["2"],["1","2"],["3"],["1","3"],["2","3"],["1","2","3"],["4"],["1","4"],["2","4"],["1","2","4"],["3","4"],["1","3","4"],["2","3","4"],["1","2","3","4"]]
--- PASS: TestArrayExhaustivity (0.00s)
PASS
Process finished with exit code 0
- 根据长度冒泡排序
func arrayLenSort(result [][]string) [][]string {
for i := 0; i < len(result); i++ {
for j := i + 1; j < len(result); j++ {
if len(result[i]) < len(result[j]) {
tmp := result[i]
result[i] = result[j]
result[j] = tmp
}
}
}
return result
}
- 示例代码3
func TestArrayExhaustivitySort(t *testing.T) {
result := getArrayExhaustivity([]string{"1", "2", "3", "4"})
result = arrayLenSort(result)
str, err := json.Marshal(result)
if err != nil {
t.Fatal(err)
}
println(string(str))
}
- 示例结果3
=== RUN TestArrayExhaustivitySort
[["1","2","3","4"],["1","2","4"],["1","3","4"],["2","3","4"],["1","2","3"],["2","4"],["1","3"],["3","4"],["2","3"],["1","2"],["1","4"],["4"],["2"],["1"],["3"]]
--- PASS: TestArrayExhaustivity (0.00s)
PASS
Process finished with exit code 0
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