题目

子集 

给定一个含不同整数的集合，返回其所有的子集

样例

如果 S = [1,2,3]，有如下的解：

[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]
注意

子集中的元素排列必须是非降序的，解集必须不包含重复的子集

挑战

你可以同时用递归与非递归的方式解决么？

解题

根据上面求排列的思想很类似，还是深度优先遍历。由于输出每个子集需要升序，所以要先对数组进行排序。求出所以的子集，也就是求出所以的组合方式 + 空集

问题转化为求组合方式的问题

参考链接不仅要考虑起始位置，还需要考虑长度，这样才是组合 C(n,k)，由于我只想到要考虑起始位置，而长度问题在程序中增加，一直没有解决问题

核心程序

    public void helper(int[] nums,int start,int len,
    ArrayList<Integer> list,ArrayList<ArrayList<Integer>> res){
        if( list.size() == len){
            res.add(new ArrayList<Integer>(list));
            return;
        }
        for(int i=start;i< nums.length;i++){
            if(list.contains(nums[i])){
                continue;
            }
            list.add(nums[i]);
            helper(nums,i+1,len,list,res);
            list.remove(list.size()-1);
            }
        
    }
class Solution {
    /**
     * @param S: A set of numbers.
     * @return: A list of lists. All valid subsets.
     */
    public ArrayList<ArrayList<Integer>> subsets(int[] nums) {
        // write your code here
        ArrayList<Integer> list = new ArrayList<Integer>();
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        if(nums == null || nums.length ==0)
            return res;
        Arrays.sort(nums);
        
        res.add(list);
        for(int len = 1;len<= nums.length;len++){
            helper(nums,0,len,list,res);
            
        }
        return res;
    }
    public void helper(int[] nums,int start,int len,
    ArrayList<Integer> list,ArrayList<ArrayList<Integer>> res){
        if( list.size() == len){
            res.add(new ArrayList<Integer>(list));
            return;
        }
        for(int i=start;i< nums.length;i++){
            if(list.contains(nums[i])){
                continue;
            }
            list.add(nums[i]);
            helper(nums,i+1,len,list,res);
            list.remove(list.size()-1);
            }
        
    }
}
Java Code

九章中程序进行了优化，长度不考虑，递归一次list的值都是子集的一个元素

class Solution {
    /**
     * @param S: A set of numbers.
     * @return: A list of lists. All valid subsets.
     */
    public ArrayList<ArrayList<Integer>> subsets(int[] nums) {
        // write your code here
        ArrayList<Integer> list = new ArrayList<Integer>();
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        if(nums == null || nums.length ==0)
            return res;
        Arrays.sort(nums);
        helper(nums,0,list,res);
     
        return res;
    }
    public void helper(int[] nums,int start, ArrayList<Integer> list,ArrayList<ArrayList<Integer>> res){
        res.add(new ArrayList<Integer>(list));
        for(int i=start;i< nums.length;i++){
            if(list.contains(nums[i])){
                continue;
            }
            list.add(nums[i]);
            helper(nums,i+1,list,res);
            list.remove(list.size()-1);
            }
        
    }
}
Java Code
class Solution:
    """
    @param S: The set of numbers.
    @return: A list of lists. See example.
    """
    def subsets(self, S):
        def dfs(depth, start, valuelist):
            res.append(valuelist)
            if depth == len(S): return
            for i in range(start, len(S)):
                dfs(depth+1, i+1, valuelist+[S[i]])
        S.sort()
        res = []
        dfs(0, 0, [])
        return res
Python Code

根据位运算进行求解

        // 1 << n is 2^n
        // each subset equals to an binary integer between 0 .. 2^n - 1
        // 0 -> 000 -> []
        // 1 -> 001 -> [1]
        // 2 -> 010 -> [2]
        // ..
        // 7 -> 111 -> [1,2,3]

 下面是我自己实现

class Solution {
    /**
     * @param S: A set of numbers.
     * @return: A list of lists. All valid subsets.
     */
    public ArrayList<ArrayList<Integer>> subsets(int[] nums) {
        // write your code here
        int len = nums.length;
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        if(nums == null || len==0)
            return res;
        Arrays.sort(nums);
        // 1 << n is 2^n
        // each subset equals to an binary integer between 0 .. 2^n - 1
        // 0 -> 000 -> []
        // 1 -> 001 -> [1]
        // 2 -> 010 -> [2]
        // ..
        // 7 -> 111 -> [1,2,3]
        for(int i=0;i< 1<<len ;i++){
            ArrayList<Integer> list = new ArrayList<Integer>();
            // 检测哪一位是 1 
            int n = i;
            for(int j=0;j< len;j++){
                if(n%2==1)
                    list.add(nums[j]);
                n=n/2;
            }
            res.add(list);
        }
        return res;
    }
}
View Code

九章中判断第几位是1的程序如下：

 for (int i = 0; i < (1 << n); i++) {
            ArrayList<Integer> subset = new ArrayList<Integer>();
            for (int j = 0; j < n; j++) {
                // check whether the jth digit in i's binary representation is 1
                if ((i & (1 << j)) != 0) {
                    subset.add(nums[j]);
                }
            }
            result.add(subset);
        }

是懂非懂，好像这样也可以判断第几位是1

class Solution:
    """
    @param S: The set of numbers.
    @return: A list of lists. See example.
    """
    def subsets(self, nums):
        # write your code here
        res = []
        size = len(nums)
        nums.sort()
        if nums == None or size == 0:
            return res
        for i in range(1<<size):
            lst=[]
            n = i
            for j in range(size):
                if n%2==1:
                    lst.append(nums[j])
                n/=2
            res.append(lst)
        return res
Python Code