# 组合总和 II
给定一个数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。
candidates 中的每个数字在每个组合中只能使用一次。
说明：
- 所有数字（包括目标数）都是正整数。
- 解集不能包含重复的组合。
示例 1:
输入: candidates = [10,1,2,7,6,1,5], target = 8,
所求解集为:[[1, 7],[1, 2, 5],[2, 6],[1, 1, 6]]
示例 2:
输入: candidates = [2,5,2,1,2], target = 5,
所求解集为:[[1,2,2],[5]]
以下程序实现了这一功能，请你填补空白处内容：
```cpp
#include
using namespace std;
class Solution
{
public:
vector> combinationSum2(vector &candidates, int target)
{
vector> res;
sort(candidates.begin(), candidates.end());
dfs(candidates, 0, target, res);
return res;
}
private:
vector stack;
void dfs(vector &candidates, int start, int target, vector> &res)
{
if (target < 0)
{
return;
}
else if (target == 0)
{
res.push_back(stack);
}
else
{
int last = INT_MIN;
for (int i = start; i < candidates.size(); i++)
{
if (last != candidates[i])
{
stack.push_back(candidates[i]);
_________________________________;
stack.pop_back();
}
last = candidates[i];
}
}
}
};
```
## template
```cpp
#include
using namespace std;
class Solution
{
public:
vector> combinationSum2(vector &candidates, int target)
{
vector> res;
sort(candidates.begin(), candidates.end());
dfs(candidates, 0, target, res);
return res;
}
private:
vector stack;
void dfs(vector &candidates, int start, int target, vector> &res)
{
if (target < 0)
{
return;
}
else if (target == 0)
{
res.push_back(stack);
}
else
{
int last = INT_MIN;
for (int i = start; i < candidates.size(); i++)
{
if (last != candidates[i])
{
stack.push_back(candidates[i]);
dfs(candidates, i + 1, target - candidates[i], res);
stack.pop_back();
}
last = candidates[i];
}
}
}
};
```
## 答案
```cpp
dfs(candidates, i + 1, target - candidates[i], res);
```
## 选项
### A
```cpp
dfs(candidates, i - 1, target - candidates[i], res);
```
### B
```cpp
dfs(candidates, i, target + candidates[i], res);
```
### C
```cpp
dfs(candidates, i + 1, target + candidates[i], res);
```