- 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 有疑问加站长微信联系（非本文作者）
