遞歸

題目：

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.

Note:

• All numbers (including target) will be positive integers.
• 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]
]

題目解析：此題要求我們在向量集合中求出元素相加等於所給目標數的組合，屬於中等難度的題目；在這裏我們首先將集合進行排序，再寫一個判斷組合的函數，用一個for循環從最小的數開始判斷，循環條件要注意i不能大於向量大小以及第i個元素不能大於目標數；在循環中調用遞歸遍歷所有的元素，之後再回溯即可。

程序如下：

class Solution {
public:
        vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int> > res;
        vector<int> combination;
        combinationSum(candidates, target, res, combination, 0);
        return res;
    }
private:
    void combinationSum(vector<int> &candidates, int target, vector<vector<int> > &res, vector<int> &combination, int begin) {
        if (!target) {
            res.push_back(combination);
            return;
        }
        for (int i = begin; i != candidates.size() && target >= candidates[i]; ++i) {
            combination.push_back(candidates[i]);
            combinationSum(candidates, target - candidates[i], res, combination, i);
            combination.pop_back();
        }
    }
};