description:
给定target， 求给定数列中找到几个数（其中的数不可以重复使用，且一组数有几个也不做限制）的和为target，和上面那个题一毛一样的，就改一下下标就行了，背下来背下来背下来~~~
Note:
All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations. //这个条件注意了，只要是两个挨着的一样，那么就会导致最后出现相同的组合Example:
Example 1:
Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]
Example 2:
Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
[1,2,2],
[5]
]answer:
class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<vector<int>> res;
vector<int> out;
sort(candidates.begin(), candidates.end()); //这里刚开始就没写，一定要先变成有序数组！！！
combinationDFS(candidates, res, out, target, 0);
return res;
}
void combinationDFS(vector<int>& candidates, vector<vector<int>>& res, vector<int>& out, int target, int start) {
if(target < 0) return;
if(target == 0) {
res.push_back(out);
return;
}
for(int i = start; i < candidates.size(); i++) {
if(i > start && candidates[i] == candidates[i - 1]) continue; //这里也要注意了，因为上边那个note的第二条
out.push_back(candidates[i]);
combinationDFS(candidates, res, out, target - candidates[i], i + 1);
out.pop_back();
}
}
};