算法精粹(algorithm-essentials)

感谢soulmachine@github提供内容
## Combination Sum


### 描述

Given a set of candidate numbers (`C`) 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.
* Elements in a combination ($$a_1, a_2, ..., a_k$$) must be in non-descending order. (ie, $$a_1 \leq a_2 \leq ... \leq a_k$$).
* 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]
```


### 分析




### 代码

{% if book.java %}
```java
// Combination Sum
// 时间复杂度O(n!),空间复杂度O(n)
public class Solution {
public List> combinationSum(int[] nums, int target) {
Arrays.sort(nums);
List> result = new ArrayList<>(); // 最终结果
List path = new ArrayList<>(); // 中间结果
dfs(nums, path, result, target, 0);
return result;
}

private static void dfs(int[] nums, List path,
List> result, int gap, int start) {
if (gap == 0) { // 找到一个合法解
result.add(new ArrayList(path));
return;
}
for (int i = start; i < nums.length; i++) { // 扩展状态
if (gap < nums[i]) return; // 剪枝

path.add(nums[i]); // 执行扩展动作
dfs(nums, path, result, gap - nums[i], i);
path.remove(path.size() - 1); // 撤销动作
}
}
}
```
{% endif %}

{% if book.cpp %}
```cpp
// Combination Sum
// 时间复杂度O(n!),空间复杂度O(n)
class Solution {
public:
vector > combinationSum(vector &nums, int target) {
sort(nums.begin(), nums.end());
vector > result; // 最终结果
vector path; // 中间结果
dfs(nums, path, result, target, 0);
return result;
}

private:
void dfs(vector& nums, vector& path, vector > &result,
int gap, int start) {
if (gap == 0) { // 找到一个合法解
result.push_back(path);
return;
}
for (size_t i = start; i < nums.size(); i++) { // 扩展状态
if (gap < nums[i]) return; // 剪枝

path.push_back(nums[i]); // 执行扩展动作
dfs(nums, path, result, gap - nums[i], i);
path.pop_back(); // 撤销动作
}
}
};
```
{% endif %}


### 相关题目

* [Combination Sum II](combination-sum-ii.md)
* [Combination Sum III](combination-sum-iii.md)