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, 0, 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 + 1, path, ret);
            path.remove(path.size() - 1);
        }
    }
}

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) {
        // 2135
        List<List<Integer>> ret = new ArrayList<List<Integer>>();
        if (S == null) {
            return ret;
        }
        
        Arrays.sort(S);
        return dfs3(S, 0, 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 + 1, 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 == 0) {
            return ret;
        }
        
        int len = S.length;
        Arrays.sort(S);
        
        // forget to add (long).
        long numOfSet = (long)Math.pow(2, len);
        
        for (int i = 0; i < numOfSet; i++) {
            // bug 3: should use tmp - i.
            long tmp = i;
            
            ArrayList<Integer> list = new ArrayList<Integer>();
            while (tmp != 0) {
                // bug 2: use error NumberOfTrailingZeros. 
                int indexOfLast1 = Long.numberOfTrailingZeros(tmp);
                list.add(S[indexOfLast1]);
                
                // clear the bit.
                tmp ^= (1 << 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