题目：

Given a collection of integers that might contain duplicates, 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 = [1,2,2], a solution is:

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

代码如下：

   void getsubset(set<vector<int> > &subsets,vector<int> &S,int i)
    {
        vector<int> tmp;
        for(int j=0;j<S.size();j++)
        {
            if(i&(1<<j))tmp.push_back(S[j]);
        }
        subsets.insert(tmp);
    }
    vector<vector<int> > subsetsWithDup(vector<int> &S) {
        set<vector<int> > subsets;
        set<vector<int> >::iterator j;
        vector<vector<int> > result;
        sort(S.begin(),S.end());
        int n=S.size();
        for(int i=0;i<1<<n;i++)
        {
            getsubset(subsets,S,i);
        }
        for(j=subsets.begin();j!=subsets.end();j++)
        {
            result.push_back(*j);
        }
        return result;
    }

题目：

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 < = m < = n < = length of list.

代码如下：

ListNode *reverseBetween(ListNode *head, int m, int n) {
        if(m==n)return head;
        int i=1;
        if(m==1)
        {
            ListNode *head1=head;
            while(i<n)
            {
                ListNode *tmp=head1->next;
                head1->next=tmp->next;
                tmp->next=head;
                head=tmp;
                i++;
            }
        }
        else
        {
            ListNode *head1=head;
            while(i<m-1)
            {
                head1=head1->next;
                i++;
            }
            ListNode *head2=head1->next;
            while(i<n-1)
            {
                ListNode *tmp=head2->next;
                head2->next=tmp->next;
                tmp->next=head1->next;
                head1->next=tmp;
                i++;
            }
        }
        return head;
    }

题目：

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,3,2].

代码如下：

    void getresult(TreeNode *root,vector<int> &result)
    {
        if(root==NULL)return;
        if(root->left!=NULL)
        {
            getresult(root->left,result);
        }
        result.push_back(root->val);
        if(root->right!=NULL)
        {
            getresult(root->right,result);
        }
        return;
    }
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> result;
        getresult(root,result);
        return result;     
    }