Subsets

Subsets

Given a set of distinct integers, nums, return all possible subsets.

Note: The solution set must not contain duplicate subsets.

For example,
If nums = [1,2,3], a solution is:

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

解析：三种方法都是听经典的方法，递归，迭代，二进制，这道题应该很快写出来的，sad...,对于n个数有2的n次方个结果，n+1的话就把原来的数再加上新的数，之前的保持不变，变成了原来的二倍。

代码：

class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        vector<vector<int>>ans;
        vector<int>path;
        subset(ans,nums,0,path);
        return ans;
    }
    
    
    void subset(vector<vector<int>>&ans,vector<int>nums,int begin,vector<int>&path)
    {
        
        ans.push_back(path);
        for (int i=begin; i<nums.size(); i++)
        {
            path.push_back(nums[i]);
            subset(ans,nums,i+1,path);
            path.pop_back();
        }
        return ;
        
    }
};

class Solution {
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        vector<vector<int>>ans;
        vector<int>path;
        ans.push_back(path);
        for (int i=0; i<nums.size(); i++)
        {
            int cnt=ans.size();
            for (int j=0; j<cnt; j++)
            {
                path=ans[j];
                path.push_back(nums[i]);
                ans.push_back(path);
            }
        }
        return ans;
    }
    
};

Subsets II

Given a collection of integers that might contain duplicates, nums, return all possible subsets.

Note: The solution set must not contain duplicate subsets.

For example,
If nums = [1,2,2], a solution is:

[
  [2],
  [1],
  [1,2,2],
  [2,2],
  [1,2],
  []
]

解析：

由于数组中可能存在重复元素，首先把数组进行排序，在求子数组时，如果当前元素与前一个元素相同则跳过当前元素。

代码：

class Solution {
public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        
        vector<vector<int>>ans;
        vector<int>path;
        int begin=0;
        sort(nums.begin(),nums.end());
        dfs(nums,ans,path,begin);
        return ans;
    }
    void dfs(vector<int>&nums,vector<vector<int>>&ans,vector<int>&path,int begin)
    {
        ans.push_back(path);
        int temp=0;
        for (int i=begin; i<nums.size(); i++)
        {
            if ((i!=begin)&&nums[i]==temp)
            continue;
            path.push_back(nums[i]);
            dfs(nums,ans,path,i+1);
            temp=path.back();
            path.pop_back();
        }
        return ;
    }
    
    
    
};