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Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:

• All numbers (including target) will be positive integers.
• Elements in a combination (a₁, a₂, … , a_k) must be in non-descending order. (ie, a₁ ≤ a₂ ≤ … ≤ a_k).
• 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]

这个题目是个backtrack的问题，这个问题，也类似于printtree中path等于固定值的解法。

其中最有意思的一个trick是 记录访问过路径的index[] 以及那个对应的position。我的第一想法是直接用一个vector来记录路径，进来一个push，回退一个popback。

最后发现这是个很笨拙的方法。最好的办法是，用一个数组index 来记录位置，用递归的层数来做索引，将简单很多。

具体这个题目，两种思路，一个是sum的方法，从sum=0 开始，把整个数组内容递归的加一遍，知道发现target，那就把路径输出。

另外一个办法是减法，

void Find(vector<int> &candidates, int sum, int target, int* index, int start)
{
    int size = candidates.size();
    if (sum>target) return;

    if (sum == target)
    {
        vector<int> result;
        for (int i = 1; i<= start; i++)
        {
            result.push_back(candidates[index[i]]);
        }
        sort(result.begin(), result.end());

        finalresult.push_back(result);
        return;
    }

    for (int k = index[start]; k< size; k++)
    {
        index[start + 1] = k;
        Find(candidates, sum + candidates[k], target, index, start + 1);
    }

}

vector<vector<int> > combinationSum(vector<int> &candidates, int target) {

    int index[1024];
    index[0] = 0;
    int start = 0;
    int sum = 0;
    Find(candidates, sum, target, index, start);

    return finalresult;
}

减法

vector<vector<int>> finalresult;
void getcombine(vector<int>candidates, int target, int* index, int start)
{
    int len = candidates.size();
    

    if (target<candidates[0])//the target not fall inside means the sequence is wrong.
    {
        return;
    }
    else 
    {
        for (int i = index[start]; i < len; i++)
        {
            index[start + 1] = i;
            if (candidates[i] == target)// find it
            {
                vector<int> result;
                //get output from index
                for (int n = 1; n <= start+1; n++)
                {
                    result.push_back(candidates[index[n]]);
                }
                finalresult.push_back(result);
                return;
            }
            getcombine(candidates, target - candidates[i], index, start+1);
        }

    }


}
vector<vector<int> > combinationSumnew(vector<int> &candidates, int target) {
    vector<int> result;
    int index[1024];
    index[0] = 0;
    int start = 0;
    
    getcombine(candidates, target, index, start);
    
    return finalresult;
}