LeetCode: 39. Combination Sum 的进阶。元素不在唯一且每一个元素只能出现一次。对结果进行排序然后通过 set 对结果进行去重：

class Solution {
public:
	vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
		sort(candidates.begin(), candidates.end());
		backtracking(candidates, 0, 0, target);
		vector<vector<int>> r;
		for (auto t : res) {
			r.push_back(t);
		}
		return  r;
	}
private:
	set<vector<int>> res;
	vector<int> track;
	map<int, bool> visited;

	void backtracking(vector<int>& condidates, int start, int n, int target) {
		if (n == target) {
			res.insert(track);
			return;
		}
		if (n > target) {
			return;
		}

		int c = 0;
		int sz = condidates.size();

		for (int i = start; i < sz; i++) {
			c = condidates[i];
			track.push_back(c);
			backtracking(condidates, i + 1, n + c, target);
			track.pop_back();
		}
	}
};

以下测试用例无法通过：

[1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]
30

class Solution {
public:
	vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
		sort(candidates.begin(), candidates.end());
		backtracking(candidates, 0, 0, target);
		vector<vector<int>> r;
		for (auto t : res) {
			r.push_back(t);
		}
		return  r;
	}
private:
	set<vector<int>> res;
	vector<int> track;
	map<int, bool> visited;

	void backtracking(vector<int>& condidates, int start, int n, int target) {
		if (n >= target) {
			if (n == target) {
				res.insert(track);
			}
			return;
		}

		int c = 0;
		int sz = condidates.size();

		for (int i = start; i < sz; i++) {
			// 避免全是一样的陷入无限的循环。
			if (i > start && condidates[i] == condidates[i - 1]) {
				continue;
			}
			c = condidates[i];
			track.push_back(c);
			backtracking(condidates, i + 1, n + c, target);
			track.pop_back();
		}
		cout << start << "/" << sz << ", ";
	}
};