Combination Sum

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, … , 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：这道题显然可以用递归做。将7进行分解，相当于一个多叉树。遍历数组，

判断target减去当前数，如果>=0则进行下层分解，如果小于则回溯。由于分解的数是非递减的，还需要判断当前

分解的数>=已分解的数的最大值,代码如下：

class Solution {private:    vector<vector<int> > res;    vector<int> perRes;public:    void combinationSumHelper(vector<int> candidates, int target, int pos, int left) {        if (target == 0) {            res.push_back(perRes);        }        else {            int len = candidates.size();            int i;            for(i=0; i<len; ++i) {                perRes.resize(pos+1);                if (target-candidates[i] >= 0 && candidates[i] >= left) {                    perRes[pos] = candidates[i];                    combinationSumHelper(candidates, target-candidates[i], pos+1,candidates[i]);                }            }        }    }    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {        res.clear();        if (candidates.empty()) {            return res;        }        perRes.clear();        combinationSumHelper(candidates,target,0,0);        return res;    }};