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# 39. Combination Sum

## problem description

Given a **set** of candidate numbers (`candidates`) **(without duplicates)** and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sums to `target`.

The **same** repeated number may be chosen from `candidates` unlimited number of times.

**Note:**

* All numbers (including `target`) will be positive integers.
* The solution set must not contain duplicate combinations.

**Example 1:**

```
Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]
```

**Example 2:**

```
Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]
```

## algorithm thought

组合数相加最后和为target，基本需要遍历所有集合。使用回溯法得到所有的集合，判断和是否为target。由于数字相加如果已经大于target，就没必要继续遍历了。所以加入了剪枝策略，首先sort数组，为了先遍历小数字后遍历大数字。然后在进入下一个函数之前判断是否满足剪枝策略，如果满足，就不进入函数。

## code

```cpp
class Solution {
public:
    std::vector<std::vector<int> > combinationSum(std::vector<int> &candidates, int target) {
         std::sort(candidates.begin(), candidates.end());
        std::vector<std::vector<int> > res;
        std::vector<int> combination;
        combinationSum(candidates, target, res, combination, 0);
        return res;
    }
private:
    void combinationSum(std::vector<int> &candidates, int target, std::vector<std::vector<int> > &res, std::vector<int> &combination, int begin) {
        if (!target) {
            res.push_back(combination);
            return;
        }
        for (int i = begin; i != candidates.size() && target >= candidates[i]; ++i)
            if (i == begin || candidates[i] != candidates[i - 1]) {
                combination.push_back(candidates[i]);
                combinationSum(candidates, target - candidates[i], res, combination, i);
                combination.pop_back();
            }
    }
};
```

## algorithm analysis

回溯法世家复杂度是O(2^n),很慢，但是这里加入了剪枝策略，还是一定程度上加快了运行速度的。