# 39. Combination Sum

## 13 Jan 39. Combination Sum

Given an array of distinct integers `candidates` and a target integer `target`, return a list of all unique combinations of `candidates` where the chosen numbers sum to `target`. You may return the combinations in any order.

The same number may be chosen from `candidates` an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.

The test cases are generated such that the number of unique combinations that sum up to `target` is less than `150` combinations for the given input.

Example 1:

```Input: candidates = [2,3,6,7], target = 7
Output: [[2,2,3],[7]]
Explanation:
2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times.
7 is a candidate, and 7 = 7.
These are the only two combinations.
```

Example 2:

```Input: candidates = [2,3,5], target = 8
Output: [[2,2,2,2],[2,3,3],[3,5]]
```

Example 3:

```Input: candidates = [2], target = 1
Output: []
```

Constraints:

• `1 <= candidates.length <= 30`
• `2 <= candidates[i] <= 40`
• All elements of `candidates` are distinct.
• `1 <= target <= 40`

Python代码如下：

class Solution(object):
def combinationSum(self, candidates, target):
“””
:type candidates: List[int]
:type target: int
:rtype: List[List[int]]
“””
res = []
candidates.sort()
self.dfs(candidates, target, 0, res, [])
return res

def dfs(self, nums, target, index, res, path):
if target < 0:
return
elif target == 0:
res.append(path)
return
for i in xrange(index, len(nums)):
if nums[index] > target:
return
self.dfs(nums, target – nums[i], i, res, path + [nums[i]])
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C++代码如下：

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

class Solution:
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
results = []
candidates.sort()
self.dfs(candidates, target, 0, [], results)
return results
def dfs(self, nums, target, index, path, results):
if target < 0 :
return
if target == 0:
results.append(path)
return
for i in range(index, len(nums)):
if nums[i] > target:
return
self.dfs(nums, target – nums[i], i, path + [nums[i]], results)