【Subsets II】cpp

题目：

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],
  []
]

代码：

class Solution {
public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
            std::sort(nums.begin(), nums.end());
            vector<vector<int> > ret;
            vector<int> tmp;
            Solution::dfs(ret, nums, 0, tmp);
            return ret;
    }
    static void dfs(vector<vector<int> >& ret, vector<int>& nums, int index, vector<int>& tmp )
    {
            if ( index==nums.size() )
            {
                ret.push_back(tmp);
                return;
            }
            tmp.push_back(nums[index]);
            Solution::dfs(ret, nums, index+1, tmp);
            tmp.pop_back();
            if ( !(tmp.size()>=1 && tmp.back()==nums[index]) )
            {
                Solution::dfs(ret, nums, index+1, tmp);
            }
    }
};

tips：

大体思路与Subsets类似（http://www.cnblogs.com/xbf9xbf/p/4516802.html）

经过画图推演：dfs向左边走的条件不变；再向右边走时，如果当前节点的最后一个元素与即将要加入的新元素相同，则不往右边走了。因为再往右边走，右边肯定包含左边的重复子集了。

===================================

再补充一个增量构造法的迭代解法，代码如下：

class Solution {
public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
            std::sort(nums.begin(), nums.end());
            vector<vector<int> > ret;
            vector<int> none;
            ret.push_back(none);
            int pureNew = 0;
            for ( size_t i = 0; i < nums.size(); ++i )
            {
                vector<vector<int> > tmp = ret;
                size_t begin = 0;
                if ( i>=1 && nums[i]==nums[i-1] )
                {
                    begin = ret.size()-pureNew;
                }
                for ( size_t j = begin; j < tmp.size(); ++j )
                {
                    tmp[j].push_back(nums[i]);
                    ret.push_back(tmp[j]);
                }
                pureNew = ret.size() - tmp.size();
            }
            return ret;
    }
};

tips:

增加新元素前，判断nums[i]==nums[i-1]的条件是否成立；如果成立，则增量应该只作用于上一轮新增加的子集上，这样保证没有重复的；这个思路也很简洁。

学习了几个blog的思路：

http://bangbingsyb.blogspot.sg/2014/11/leetcode-subsets-i-ii.html

http://www.cnblogs.com/TenosDoIt/p/3451902.html

http://www.cnblogs.com/yuzhangcmu/p/4211815.html

完毕。

============================================

第二次过这道题，只用dfs的写法。再次画画图，理解一下。

class Solution {
public:
        vector<vector<int> > subsetsWithDup(vector<int>& nums)
        {
            sort(nums.begin(), nums.end());
            vector<vector<int> > ret;
            vector<int> tmp;
            Solution::dfs(ret, nums, 0, tmp);
            return ret;
        }
        static void dfs(
            vector<vector<int> >& ret, 
            vector<int>& nums, int index, 
            vector<int>& tmp)
        {
            if ( index==nums.size() )
            {
                ret.push_back(tmp);
                return;
            }
            tmp.push_back(nums[index]);
            Solution::dfs(ret, nums, index+1, tmp);
            tmp.pop_back();
            if ( !(tmp.size()>0 && tmp.back()==nums[index]) )
            {
                Solution::dfs(ret, nums, index+1, tmp);
            }
        }
};