java求解集合的子集的实例

@Test 
public void test1() throws Exception { 
   
  Set> subsets = getSubsets( Arrays.asList(1,2,6)); 
  Set> subsets2 = getSubsets( Arrays.asList("a","b","c")); 
  Set> subsets3 = getSubsets( Arrays.asList('b','c','d')); 
  System.out.println(subsets); 
  System.out.println(subsets2); 
  System.out.println(subsets3); 
} 
 
//集合接受各种类型数据 
public  Set> getSubsets(List subList) { 
  //考虑去重 
  Set> allsubsets = new LinkedHashSet<>(); 
  int max = 1 << subList.size(); 
  for (int loop = 0; loop < max; loop++) { 
    int index = 0; 
    int temp = loop; 
    ArrayList  currentCharList = new ArrayList(); 
    //控制索引 
    while (temp > 0) { 
      if ((temp & 1) > 0) { 
        currentCharList.add(subList.get(index)); 
      } 
      temp >>= 1; 
      index++; 
    } 
    allsubsets.add(currentCharList); 
  } 
  return allsubsets; 
} 

   @Test 
public void testName() throws Exception { 
  Set> subsets2 = getSubsets2(Arrays.asList(1,2,3)); 
  System.out.println(subsets2); 
} 
 
//方式2 归纳法 
//从{}和最后一个元素开始,每次迭代加一个元素组成一个新的集合 
public  Set> getSubsets2(List list) { 
   if (list.isEmpty()) { 
     Set> ans=new LinkedHashSet<>(); 
     ans.add(Collections.emptyList()); 
     return ans; 
   } 
    
   Integer first=list.get(0); 
   List rest=list.subList(1, list.size()); 
   Set> list1 = getSubsets2(rest); 
   Set> list2 = insertAll(first, list1);// 
   System.out.println(list1); 
   System.out.println(list2); 
   System.out.println("================"); 
   return concat(list1, list2); 
} 
 
 
   public  Set> insertAll(Integer first,Set> lists){ 
  // 
  Set> result=new LinkedHashSet<>(); 
  for (List list : lists) { 
    List copy=new ArrayList<>(); 
    copy.add(first); 
    copy.addAll(list); 
    result.add(copy); 
  } 
  return result; 
} 
   
  //这样写可以不影响lists1,lists2的值 
  private Set> concat(Set> lists1,Set> lists2) { 
  Set> temp=new LinkedHashSet<>(lists1); 
  temp.addAll(lists2); 
  return temp; 
}