LeetCode: Subsets 解题报告

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:

使用九章算法的模板：

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

 public class Solution {
     public List<List<Integer>> subsets(int[] S) {
         List<List<Integer>> ret = new ArrayList<List<Integer>>();
         if (S == null) {
             return ret;
         }

         Arrays.sort(S);

         dfs(S, , new ArrayList<Integer> (), ret);

         return ret;
     }

     public void dfs(int[] S, int index, List<Integer> path, List<List<Integer>> ret) {
         ret.add(new ArrayList<Integer>(path));

         for (int i = index; i < S.length; i++) {
             path.add(S[i]);
             dfs(S, i + , path, ret);
             path.remove(path.size() - );
         }
     }
 }

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

 // Solution 3: The memory and recursion.
     public List<List<Integer>> subsets(int[] S) {
         //
         List<List<Integer>> ret = new ArrayList<List<Integer>>();
         if (S == null) {
             return ret;
         }

         Arrays.sort(S);
         return dfs3(S, , new HashMap<Integer, List<List<Integer>>>());
     }

     public List<List<Integer>> dfs3(int[] S, int index, HashMap<Integer, List<List<Integer>>> map) {
         int len = S.length;

         if (map.containsKey(index)) {
             return map.get(index);
         }

         List<List<Integer>> ret = new ArrayList<List<Integer>>();
         List<Integer> pathTmp = new ArrayList<Integer>();
         ret.add(pathTmp);

         for (int i = index; i < len; i++) {
             List<List<Integer>> left = dfs3(S, i + , map);
             for (List<Integer> list: left) {
                 pathTmp = new ArrayList<Integer>();
                 pathTmp.add(S[i]);
                 pathTmp.addAll(list);
                 ret.add(pathTmp);
             }
         }

         map.put(index, ret);
         return ret;
     }

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: http://www.fusu.us/2013/07/the-subsets-problem.html

There are many variations of this problem, I will stay on the general problem of finding all subsets 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 combinations.

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}

 public class Solution {
     public List<List<Integer>> subsets(int[] S) {
         List<List<Integer>> ret = new ArrayList<List<Integer>>();
         if (S == null || S.length == ) {
             return ret;
         }

         int len = S.length;
         Arrays.sort(S);

         // forget to add (long).
         long numOfSet = (long)Math.pow(, len);

         for (int i = ; i < numOfSet; i++) {
             // bug 3: should use tmp - i.
             long tmp = i;

             ArrayList<Integer> list = new ArrayList<Integer>();
             while (tmp != ) {
                 // bug 2: use error NumberOfTrailingZeros.
                 int indexOfLast1 = Long.numberOfTrailingZeros(tmp);
                 list.add(S[indexOfLast1]);

                 // clear the bit.
                 tmp ^= ( << indexOfLast1);
             }

             ret.add(list);
         }

         return ret;
     }

 }

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: https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/dfs/Subsets.java