[LeetCode] Combination Sum

Note:

• All numbers (including target) will be positive integers.
• Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
• The solution set must not contain duplicate combinations.

For example, given candidate set 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3] 

思路，对于2，从2开始，对于3，从3开始，保证不会出现322，只会出现223， 其实还是回溯法，或者说dfs，如下图所示

code1：在红字出剪纸，

class Solution
{
    vector<vector<int> > m_result;
    public:

    void dfs(const vector<int> &candidates, int target, vector<int>& array, int start) 
    {   
        if(target < 0)
            return;
        if(target == 0)
        {   
            m_result.push_back(array);
            return;
        }   

        for(size_t i = start; i < candidates.size(); i++)
        {   
            array.push_back(candidates[i]);
            dfs(candidates, target - candidates[i], array, i); 
            array.pop_back();
        }   
    }   

    vector<vector<int> > combinationSum(vector<int> &candidates, int target)
    {   
        vector<int> array;
        sort(candidates.begin(), candidates.end());
        vector<int>::iterator pos = unique(candidates.begin(), candidates.end());  
        candidates.erase(pos, candidates.end()); 
        dfs( candidates, target, array, 0); 

        return m_result;
    }   
};

code2：在红字出剪纸。

class Solution
{
    vector<vector<int> > m_result;
    public:

    void dfs(const vector<int> &candidates, int target, vector<int>& array, int start) 
    {   
        if(target == 0)
        {   
            m_result.push_back(array);
            return;
        }   

        for(size_t i = start; i < candidates.size(); i++)
        {   
            if(target < candidates[i])
                return;
            array.push_back(candidates[i]);
            dfs(candidates, target - candidates[i], array, i); 
            array.pop_back();
        }   
    }   

    vector<vector<int> > combinationSum(vector<int> &candidates, int target)
    {   
        vector<int> array;
        sort(candidates.begin(), candidates.end());
        vector<int>::iterator pos = unique(candidates.begin(), candidates.end());  
        candidates.erase(pos, candidates.end()); 
        dfs( candidates, target, array, 0); 

        return m_result;
    }   
};