## Combination Sum II

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 (a1a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
• 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]

### Analysis:

The only difference between this problem and the previous one is each element can be used at most once.
We can change the recursive step to handle this.

### Code(C++):

class Solution {
public:
void dfs(vector<int> &num, int target, vector<vector<int> > &res, vector<int> &r,int st){
if (target<0){
return;
}else{
if (target==0){
res.push_back(r);
}else{
int pre = -1;
for (int i=st;i<num.size();i++){
if (num[i]!=pre){
r.push_back(num[i]);
dfs(num,target-num[i],res,r,i+1);
pre = num[i];
r.pop_back();
}
}
}
}
}

vector<vector<int> > combinationSum2(vector<int> &num, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<vector<int> > res;
if (num.size()==0){return res;}
sort(num.begin(),num.end());
vector<int> r;
dfs(num,target,res,r,0);
return res;
}
};

### Code(Python):

class Solution:
def search(self, candidates, target, i, res, ress):
if target < 0:
return
else:
if target == 0:
res.append(ress[:]) #It is important to use "ress[:]" instead of "ress"
else:
for j in xrange(i, len(candidates)):
if j==0 or candidates[j] != candidates[j-1]:
ress.append(candidates[j])
self.search(candidates, target-candidates[j], j+1, res, ress)
ress.pop(-1) #if use "ress", this will pop the elemtnes in res also

# @param {integer[]} candidates
# @param {integer} target
# @return {integer[][]}
def combinationSum2(self, candidates, target):
res =[]
ress = []
self.search(sorted(candidates), target, 0, res, ress)
return res