题目
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]
分析
与上一题39 Combination Sum本质相同，只不过需要注意两点：每个元素只能出现结果序列中一次，结果序列不可重复。
只需利用find函数添加一个判重即可。
AC代码
class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
if (candidates.empty())
return vector<vector<int> >();
sort(candidates.begin(), candidates.end());
ret.clear();
vector<int> tmp;
combination(candidates, 0, tmp, target);
return ret;
}
void combination(vector<int> &candidates, int idx, vector<int> &tmp, int target)
{
if (target == 0)
{
if (find(ret.begin(), ret.end(), tmp) == ret.end())
ret.push_back(tmp);
return;
}
else{
int size = candidates.size();
for (int i = idx; i < size; ++i)
{
if (target >= candidates[i])
{
tmp.push_back(candidates[i]);
combination(candidates, i + 1, tmp, target - candidates[i]);
tmp.pop_back();
}//if
}//for
}//else
}
private:
vector<vector<int> > ret;
};