组合算法

Question

    public List<List<Fieldmatrix>> permute(List<Fieldmatrix> fields, int start, List<Fieldmatrix> prefix) {

    List<List<Fieldmatrix>> combinations = new ArrayList<>();

        if (start >= fields.size()) {
            combinations.add(prefix);
            return combinations;
        }
        String field = fields.get(start).getFieldName();
        if (fields.get(start).isMandatory() == false) {
            combinations.addAll(combine(fields, start + 1, prefix));
        }
        List<Fieldmatrix> prefixWithField = new ArrayList<>(prefix);
        prefixWithField.add(new Fieldmatrix (field, fields.get(start).isMandatory(), fields.get(start).getFieldtype(), new ArrayList<>()));
        combinations.addAll(combine(fields, start + 1, prefixWithField));

        return combinations;
    }

• 字段1，为true
• 字段2，真
• 字段3,false
• 字段4,false
• 字段5,false

结果是：

• {Field1,Field2,Field3,Field4,Field5}
• {Field1,Field2,,Field4,Field5}
• {Field1,Field2,,,Field5}
• {Field1,Field2,,,}
• {Field1,Field2,Field3,,Field5}
• {Field1,Field2,Field3,,}
• {Field1,Field2,Field3,Field4,}
• {Field1,Field2,,Field4,}

public class Fieldmatrix {

private String fieldName;
private boolean mandatory;
private Type fieldtype;
private List<Fieldmatrix> list = new ArrayList<>();

public Fieldmatrix(String fieldName, boolean mandatory, Type fieldtype, List<Fieldmatrix> list){
    this.fieldName = fieldName;
    this.mandatory = mandatory;
    this.fieldtype = fieldtype;
    this.list = list;
}
//Getters and Setters

null

Answer 1

假设顺序不重要，您可以使用一种方法来生成可选参数（即在您的情况下为假参数）的所有组合，如下所示：

假设您有3个可选参数和2个强制参数。所以您将有3^2=9个组合。

现在，您所要做的就是迭代9次，生成所有可能的组合。

for(i=0;i<9;i++){
  for(j=0;j<options.length;j++){
    if(j'th bit is set in 'i')
      print(options[j]);
  }
}