## 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
return;
}
for (int i = start; i < nums.length; i++) { // 扩展状态
if (gap < nums[i]) return; // 剪枝

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)