地址:https://leetcode.com/problems/combination-sum-ii/

題目：

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

Each number in candidates 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.

Example 1:

Example 2:

理解：

是39 Combination Sum的後續。區別在於，本題不能重複使用元素，但是集合有重複元素。
和上一題的思想是一樣的，需要修改一些實現。
首先，不能重複使用，遞迴呼叫的時候取begin+1而不是begin就可以避免重複使用。
對於重複結果，僅當元素在集合中首次出現的時候才新增並遞迴呼叫即可。

實現：

Accepted 16ms c++ solution use backtracking, easy understand.

class Solution {
public:
	vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
		sort(candidates.begin(), candidates.end());
		vector<vector<int>> res;
		vector<int> combination;
		combinationSum(candidates, res, target, combination, 0);
		return res;
	}

	void combinationSum(vector<int>& candidates, vector<vector<int>>& res, int target, vector<int>& combination, int begin) {
		if (!target) {
			res.push_back(combination);
			return;
		}
		for (int i = begin; i < candidates.size() && target >= candidates[i]; ++i) {
			//avoid repeated res
			if (i == begin || candidates[i] != candidates[i - 1]) {
				combination.push_back(candidates[i]);
				// i+1 avoid repeatedly using element in the candidates
				combinationSum(candidates, res, target - candidates[i], combination, i + 1);
				combination.pop_back();
			}
		}
	}
};