# 小旭讲解 LeetCode 39. Combination Sum 回溯法

## 原题

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]
]```

## 讲义

回溯法（深度优先搜索，Depth-First Search）

## 代码

```class Solution {
public:
vector<vector <int>> combinationSum(vector<int>& candidates, int target) {
vector<vector <int>> result{};
vector<int> candidate{};
sort(candidates.begin(), candidates.end());
sum(result, candidates, target, candidate, 0);
return result;
}

void sum(vector<vector <int>>& result, vector<int>& candidates, int target, vector</int><int>& candidate, int level) {
// up to the target
if (target == 0) {
result.push_back(candidate);
return;
}
for (int i = level; i < candidates.size() && candidates[i] <= target; ++i) {
candidate.push_back(candidates[i]);
sum(result, candidates, target - candidates[i], candidate, i);
candidate.pop_back();
}
}
};
```