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 (a1, a2, … , 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]
标准递归回溯法可以搞定，防止重复，使用set容器。
class Solution {
public:
vector<vector<int> > combinationSum2(vector<int> &num, int target)
{
sort(num.begin(), num.end());
vector<int> tmp;
set<vector<int> > rs;
comb(rs, tmp, num, target);
return vector<vector<int> >(rs.begin(), rs.end());
}
void comb(set<vector<int> > &rs, vector<int> &tmp,
vector<int> &num, int tar, int index=0)
{
if (!tmp.empty() && tar == 0) rs.insert(tmp);
for (int i = index; i < num.size(); i++)
{
if (tar < num[i]) return;
tmp.push_back(num[i]);
comb(rs, tmp, num, tar-num[i], i+1);
tmp.pop_back();
}
}
};