39. Combination Sum

Given a set of candidate numbers (C) (without duplicates) 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.

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

给一个数组C和一个目标值T，求C中数字的各种组合，使其和等于目标T，每个元素可以使用无限次。

用回溯法穷举就可以了，不过要限定只能向后选数字，不然会造成重复。具体在算法实现中，添加一个当前判定数字位置的pos变量就可以了。

class Solution {
public:
    void helper(vector<int>& candidates, int target, int sum, int pos, vector<int> v, vector<vector<int>>& res) {
        if (sum==target) {
            res.push_back(v);
            return;
        }
        for (int i=pos; i<candidates.size(); ++i) {
            if (sum+candidates[i]<=target) {
                v.push_back(candidates[i]);
                sum += candidates[i];
                helper(candidates, target, sum, i, v, res);
                sum -= candidates[i];
                v.pop_back();
            }
        }
    }
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int>> res;
        vector<int> v;
        helper(candidates, target, 0, 0, v, res);
        return res;
    }
};