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.
- 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] ]
该题同Combination Sum类似，只不过要求结果数组中不重复
1、先对给定candidates数组排序，这样使找出的每个组合中元素排序一致。便于去重
2、使用std::find()函数对每个新组合判断，将不重复的组合放入结果数组中
3、因为题中要求每个数字只能使用一次，则helper函数中迭代时，idx递增1.
class Solution { public: vector<vector<int>> combinationSum2(vector<int>& candidates, int target) { vector<vector<int>> res; vector<int> tmp; sort(candidates.begin(), candidates.end()); int idx = 0; helper(res, tmp, candidates, target, idx); return res; } void helper(vector<vector<int>>& res, vector<int>& tmp, vector<int>& candidates, int target, int idx) { if (target < 0) { return; } else if (target == 0) { if (find(res.begin(), res.end(), tmp) == res.end()) res.push_back(tmp); } else { for (int i = idx; i < candidates.size(); i++) { tmp.push_back(candidates[i]); helper(res, tmp, candidates, target - candidates[i], i + 1); tmp.pop_back(); } } } }; // 18 ms
