78. Subsets

题目：

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],
  []
]

链接： http://leetcode.com/problems/subsets/

题解：

求数组子数组。先把数组排序，之后就可以使用DFS，维护一个递增的position，递归后要backtracking。

Time Complexity - O(n * 2ⁿ)， Space Complexity - O(n)

public class Solution {
    public ArrayList<ArrayList<Integer>> subsets(int[] S) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if(S == null || S.length == 0)
            return result;
        ArrayList<Integer> list = new ArrayList<Integer>();
        Arrays.sort(S);
        helper(result, list, S, 0);
        return result;
    }
    
    private void helper(ArrayList<ArrayList<Integer>> result, ArrayList<Integer> list, int[] S, int pos){
        result.add(new ArrayList<Integer>(list));
        
        for(int i = pos; i < S.length; i ++){
            list.add(S[i]);
            helper(result, list, S, ++pos);
            list.remove(list.size() - 1);
        }
    }
}

 

Updates:

public class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        if(nums == null || nums.length == 0)
            return res;
        Arrays.sort(nums);
        ArrayList<Integer> list = new ArrayList<>();
        dfs(res, list, nums, 0);
        return res;
    }
    
    private void dfs(List<List<Integer>> res, ArrayList<Integer> list, int[] nums, int pos) {
        res.add(new ArrayList<Integer>(list));
        
        for(int i = pos; i < nums.length; i++) {
            list.add(nums[i]);
            dfs(res, list, nums, ++pos);
            list.remove(list.size() - 1);
        }
    }
}

 

Update:

为什么以前总写成++pos？ i + 1就可以了

public class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        if(nums == null || nums.length == 0)
            return res;
        Arrays.sort(nums);
        ArrayList<Integer> list = new ArrayList<>();
        dfs(res, list, nums, 0);
        return res;
    }
    
    private void dfs(List<List<Integer>> res, ArrayList<Integer> list, int[] nums, int pos) {
        res.add(new ArrayList<Integer>(list));
        
        for(int i = pos; i < nums.length; i++) {
            list.add(nums[i]);
            dfs(res, list, nums, i + 1);
            list.remove(list.size() - 1);
        }
    }
}

 

二刷：

发现自己以前不懂装懂糊弄过去了好多题...我勒个去。

这道题目我们也是使用跟上一题combination类似的方法。

1. 这里我们根据题意首先要对数组排个序
2. 构造一个辅助函数getSubsets来进行DFS和backtracking， 同时这个辅助函数还要有一个pos来控制遍历的位置，我们先pass 0 进去。
3. 每次进入getSubsets我们都直接往结果集中加入一个当前List的新副本
4. 接下来从pos开始遍历整个数组，每次进入新一层dfs的时候pass 新的pos =  i + 1，这样就能保证结果中的顺序是从小到大

Java:

Time Complexity - O(n!), Space Complexity (n²)

public class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        if (nums == null || nums.length == 0) {
            return res;
        }
        Arrays.sort(nums);
        List<Integer> list = new ArrayList<>();
        getSubsets(res, list, nums, 0);
        return res;
    }
    
    private void getSubsets(List<List<Integer>> res, List<Integer> list, int[] nums, int pos) {
        res.add(new ArrayList<Integer>(list));
        for (int i = pos; i < nums.length; i++) {
            list.add(nums[i]);
            getSubsets(res, list, nums, i + 1);
            list.remove(list.size() - 1);
        }
    }
}

 

三刷:

下次还需要研究Bit Manipulation 以及 iterative的写法。

Java:

Time Complexity - O(n * 2ⁿ), Space Complexity (2ⁿ)

 

class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        if (nums == null || nums.length == 0) return res;
        subsets(res, nums, new ArrayList<Integer>(), 0);
        return res;
    }
    
    private void subsets(List<List<Integer>> res, int[] nums, List<Integer> list, int idx) {
        res.add(new ArrayList<>(list));
        
        for (int i = idx; i < nums.length; i++) {
            list.add(nums[i]);
            subsets(res, nums, list, i + 1);
            list.remove(list.size() - 1);
        }
    }
}

　　

 

测试：

Reference: 

https://leetcode.com/discuss/72498/simple-iteration-no-recursion-no-twiddling-explanation

https://leetcode.com/discuss/25696/simple-java-solution-with-for-each-loops

https://leetcode.com/discuss/29631/java-subsets-solution

https://leetcode.com/discuss/46668/recursive-iterative-manipulation-solutions-explanations

https://leetcode.com/discuss/9213/my-solution-using-bit-manipulation

http://www.cnblogs.com/springfor/p/3879830.html

http://www.cnblogs.com/zhuli19901106/p/3492515.html

http://www.1point3acres.com/bbs/thread-117602-1-1.html