LeetCode[Backtracking]: 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],
[]
]

我的想法是常规解法，利用递归来解，我的C++代码实现如下：

class Solution {public:    vector<vector<int> > subsets(vector<int> &S) {        vector<vector<int> > res;        sort(S.begin(), S.end());        subsets(S, -1, res);        return res;    }private:    void subsets(vector<int> &S, int start, vector<vector<int> > &res) {        if (start != S.size()) {            if (start == -1) res.push_back(vector<int> ());            else                for (int i = 0, size = res.size(); i < size; ++i) {                    vector<int> set = res[i];                    set.push_back(S[start]);                    res.push_back(set);                }            subsets(S, start + 1 ,res);        }    }};

这个discuss提出了一种令人惊喜的解法：https://leetcode.com/discuss/9213/my-solution-using-bit-manipulation

回复里面有非常好的解释：

This is an amazing solution.Learnt a lot.Let me try to explain this to those who didn’t get the logic.
Number of subsets for {1 , 2 , 3 } = 2^3 .
why ?
case possible outcomes for the set of subsets
1 -> Take or dont take = 2
2 -> Take or dont take = 2
3 -> Take or dont take = 2
therefore , total = 2*2*2 = 2^3 = { { } , {1} , {2} , {3} , {1,2} , {1,3} , {2,3} , {1,2,3} }
Lets assign bits to each outcome -> First bit to 1 , Second bit to 2 and third bit to 3
Take = 1
Dont take = 0
0) 0 0 0 -> Dont take 3 , Dont take 2 , Dont take 1 = { }
1) 0 0 1 -> Dont take 3 , Dont take 2 , take 1 = {1 }
2) 0 1 0 -> Dont take 3 , take 2 , Dont take 1 = { 2 }
3) 0 1 1 -> Dont take 3 , take 2 , take 1 = { 1 , 2 }
4) 1 0 0 -> take 3 , Dont take 2 , Dont take 1 = { 3 }
5) 1 0 1 -> take 3 , Dont take 2 , take 1 = { 1 , 3 }
6) 1 1 0 -> take 3 , take 2 , Dont take 1 = { 2 , 3 }
7) 1 1 1 -> take 3 , take 2 , take 1 = { 1 , 2 , 3 }
In the above logic ,Insert S[i] only if (j>>i)&1 ==true { j E { 0,1,2,3,4,5,6,7 } i = ith element in the input array }
element 1 is inserted only into those places where 1st bit of j is 1
if( j >> 0 &1 ) ==> for above above eg. this is true for sl.no.( j )= 1 , 3 , 5 , 7
element 2 is inserted only into those places where 2nd bit of j is 1
if( j >> 1 &1 ) == for above above eg. this is true for sl.no.( j ) = 2 , 3 , 6 , 7
element 3 is inserted only into those places where 3rd bit of j is 1
if( j >> 2 & 1 ) == for above above eg. this is true for sl.no.( j ) = 4 , 5 , 6 , 7
Time complexity : O(n*2^n) , for every input element loop traverses the whole solution set length i.e. 2^n

C++代码实现如下：

    vector<vector<int> > subsets(vector<int> &S) {        int set_num = pow(2, S.size());        vector<vector<int> > res(set_num);        sort(S.begin(), S.end());        for (int i = 0; i < S.size(); ++i)            for (int j = 0; j < set_num; ++j)                if (j & (1 << i)) res[j].push_back(S[i]);        return res;    }

这种方法的思维令人惊艳！