【leetcode】Subsets 1&2

Subsets

Given a set of distinct integers, nums, 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 nums = [1,2,3], a solution is:
[ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], [] ]
思路：
深度搜索，元素存在或者不存在两种可能，然后进行深搜。

class Solution {public:    vector<vector<int> > res;    vector<int> temp;    void sub(vector<int>& nums, int index)    {        if(index==nums.size())        {            res.push_back(temp);            return;        }        sub(nums,index+1);        temp.push_back(nums[index]);        sub(nums,index+1);        temp.pop_back();    }    vector<vector<int> > subsets(vector<int>& nums) {        sort(nums.begin(),nums.end());        sub(nums,0);               return res;    }};

Subsets II

Given a collection of integers that might contain duplicates, nums, 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 nums = [1,2,2], a solution is:
[ [2], [1], [1,2,2], [2,2], [1,2], [] ]
思路：
(1) 用STL的库函数，对vector的数去重unique(res.begin(),res.end());
(2) 在subset问题的基础上，修改，在push_back的基础上，判断是否一致。

class Solution {public:    vector<vector<int> > res;    vector<int> temp;    void sub(vector<int>& nums, int index)    {        if(index==nums.size())        {            for(int i=0;i<res.size();i++)            {                if(temp==res[i])                    return;            }            res.push_back(temp);            return;        }        sub(nums,index+1);        temp.push_back(nums[index]);        sub(nums,index+1);        temp.pop_back();    }    vector<vector<int> > subsetsWithDup(vector<int>& nums) {        sort(nums.begin(),nums.end());        sub(nums,0);        //unique(res.begin(),res.end());        return res;    }};