## Combination Sum II

### 描述

Given a collection of candidate numbers (`C`) and a target number (`T`), find all unique combinations in `C` where the candidate numbers sums to `T`.

Each number in `C` may only be used **once** in the combination.

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 > a_2 > ... > a_k\$\$).
* The solution set must not contain duplicate combinations.

For example, given candidate set `10,1,2,7,6,1,5` and target `8`,
A solution set is:

```
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]
```

### 分析

### 代码

{% if book.java %}
```java
// Combination Sum II
// 时间复杂度O(n!)，空间复杂度O(n)
public class Solution {
public List> combinationSum2(int[] nums, int target) {
Arrays.sort(nums); // 跟第 50 行配合，
// 确保每个元素最多只用一次
List> result = new ArrayList<>();
List path = new ArrayList<>();
dfs(nums, path, result, target, 0);
return result;
}
// 使用nums[start, nums.size())之间的元素，能找到的所有可行解
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;
}

int previous = -1;
for (int i = start; i < nums.length; i++) {
// 如果上一轮循环已经使用了nums[i]，则本次循环就不能再选nums[i]，
// 确保nums[i]最多只用一次
if (previous == nums[i]) continue;

if (gap < nums[i]) return; // 剪枝

previous = nums[i];

dfs(nums, path, result, gap - nums[i], i + 1);
path.remove(path.size() - 1); // 恢复环境
}
}
}
```
{% endif %}

{% if book.cpp %}
```cpp
// Combination Sum II
// 时间复杂度O(n!)，空间复杂度O(n)
class Solution {
public:
vector > combinationSum2(vector &nums, int target) {
sort(nums.begin(), nums.end()); // 跟第 50 行配合，
// 确保每个元素最多只用一次
vector > result;
vector path;
dfs(nums, path, result, target, 0);
return result;
}
private:
// 使用nums[start, nums.size())之间的元素，能找到的所有可行解
static void dfs(const 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;
}

int previous = -1;
for (size_t i = start; i < nums.size(); i++) {
// 如果上一轮循环已经使用了nums[i]，则本次循环就不能再选nums[i]，
// 确保nums[i]最多只用一次
if (previous == nums[i]) continue;

if (gap < nums[i]) return; // 剪枝

previous = nums[i];

path.push_back(nums[i]);
dfs(nums, path, result, gap - nums[i], i + 1);
path.pop_back(); // 恢复环境
}
}
};
```
{% endif %}

### 相关题目

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