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