Subsets

Given a set of distinct integers, S, return all possible subsets.

Note:

• Elements in a subset must be in non-descending order.
• The solution set must not contain duplicate subsets.

 

For example,

If S = [1,2,3], a solution is:

[

  [3],

  [1],

  [2],

  [1,2,3],

  [1,3],

  [2,3],

  [1,2],

  []

]

SOLUTION 1:

使用的模板：

递归解决。

1. 先对数组进行排序。

2. 在set中依次取一个数字出来即可，因为我们保持升序，所以不需要取当前Index之前的数字。

TIME: 227 ms

1 public class Solution { 2     public List
      
       
        > subsets(int[] S) { 3         List
        
         
          > ret = new ArrayList
          
           
            >(); 4 if (S == null) { 5 return ret; 6 } 7 8 Arrays.sort(S); 9 10 dfs(S, 0, new ArrayList
            
              (), ret);11 12 return ret;13 }14 15 public void dfs(int[] S, int index, List
             
               path, List
              
               
                > ret) {16 ret.add(new ArrayList
                
                 (path));17 18 for (int i = index; i < S.length; i++) {19 path.add(S[i]);20 dfs(S, i + 1, path, ret);21 path.remove(path.size() - 1);22 }23 }24 }
                
               
              
             
            
           
          
         
        
       
      

SOLUTION 2:

在Solution 1的基础之上，使用Hashmap来记录中间结果，即是以index开始的所有的组合，希望可以加快运行效率，最后时间：

TIME:253 ms.

实际结果与预期反而不一致。原因可能是每次新组装这些解也需要耗费时间

1 // Solution 3: The memory and recursion. 2     public List
     
      
       > subsets(int[] S) { 3         // 2135 4         List
       
        
         > ret = new ArrayList
         
          
           >(); 5 if (S == null) { 6 return ret; 7 } 8 9 Arrays.sort(S);10 return dfs3(S, 0, new HashMap
           
            >>());11 }12 13 public List
            
             
              > dfs3(int[] S, int index, HashMap
              
               >> map) {14 int len = S.length;15 16 if (map.containsKey(index)) {17 return map.get(index);18 }19 20 List
               
                
                 > ret = new ArrayList
                 
                  
                   >();21 List
                   
                     pathTmp = new ArrayList
                    
                     ();22 ret.add(pathTmp);23 24 for (int i = index; i < len; i++) {25 List
                     
                      
                       > left = dfs3(S, i + 1, map);26 for (List
                       
                         list: left) {27 pathTmp = new ArrayList
                        
                         ();28 pathTmp.add(S[i]);29 pathTmp.addAll(list);30 ret.add(pathTmp);31 }32 }33 34 map.put(index, ret);35 return ret;36 }
                        
                       
                      
                     
                    
                   
                  
                 
                
               
              
             
            
           
          
         
        
       
      
     

 

SOLUTION 3:

相当牛逼的bit解法。基本的想法是，用bit位来表示这一位的number要不要取，第一位有1，0即取和不取2种可能性。所以只要把0到N种可能

都用bit位表示，再把它转化为数字集合，就可以了。

Ref: 

There are many variations of this problem, I will stay on the general problem of finding all  of a set. For example if our set is [1, 2, 3] - we would have 8 (2 to the power of 3) subsets: {[], [1], [2], [3], [1, 2], [1, 3], [1, 2, 3], [2, 3]}. So basically our algorithm can't be faster than O(2^n) since we need to go through all possible .

 

There's a few ways of doing this. I'll mention two ways here - the recursive way, that we've been taught in high schools; and using a bit string.

 

Using a bit string involves some bit manipulation but the final code can be found easy to understand. The idea  is that all the numbers from 0 to 2^n are represented by unique bit strings of n bit width that can be translated into a subset. So for example in the above mentioned array we would have 8 numbers from 0 to 7 inclusive that would have a bit representation that is translated using the bit index as the index of the array.

 

Nr	Bits	Combination
0	000	{}
1	001	{1}
2	010	{2}
3	011	{1, 2}
4	100	{3}
5	101	{1, 3}
6	110	{2, 3}
7	111	{1, 2, 3}

1 public class Solution { 2     public List
      
       
        > subsets(int[] S) { 3         List
        
         
          > ret = new ArrayList
          
           
            >(); 4 if (S == null || S.length == 0) { 5 return ret; 6 } 7 8 int len = S.length; 9 Arrays.sort(S);10 11 // forget to add (long).12 long numOfSet = (long)Math.pow(2, len);13 14 for (int i = 0; i < numOfSet; i++) {15 // bug 3: should use tmp - i.16 long tmp = i;17 18 ArrayList
            
              list = new ArrayList
             
              ();19 while (tmp != 0) {20 // bug 2: use error NumberOfTrailingZeros. 21 int indexOfLast1 = Long.numberOfTrailingZeros(tmp);22 list.add(S[indexOfLast1]);23 24 // clear the bit.25 tmp ^= (1 << indexOfLast1);26 }27 28 ret.add(list);29 }30 31 return ret;32 }33 34 }
             
            
           
          
         
        
       
      

 性能测试：

1. when SIZE = 19:

Subset with memory record: 14350.0 millisec.

Subset recursion: 2525.0 millisec.

Subset Iterator: 5207.0 millisec.

表明带memeory的性能反而不行。而iterator的性能也不并不如。

 

2. size再继续加大时，iterator的会出现Heap 溢出的问题，且速度非常非常慢。原因不是太懂。

GITHUB: