Leetcode dfs Combination SumII

Combination Sum II

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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] 

题意：给定一组数C和一个数值T，在C中找到全部总和等于T的组合。

C中的同一数字最多仅仅能拿一次。找到的组合不能反复。
思路：dfs
第i层的第j个节点有  n - i - j 个选择分支
递归深度：递归到总和大于等于T就能够返回了
复杂度：时间O(n!)，空间O(n)

vector<vector<int> > res;
vector<int> _num;
void dfs(int start, int target, vector<int> &path){
	if(target == 0) {res.push_back(path); return;}
	int previous = -1; //这里要加上这个来记录同一层分枝的前一个值。假设当前值跟前一个值一样。就跳过，避免反复
	for(int i = start; i < _num.size(); ++i){
		if(previous == _num[i]) continue;
		if(target < _num[i]) return; //剪枝
		previous = _num[i];
		path.push_back(_num[i]);
		dfs(i + 1, target - _num[i], path);
		path.pop_back();
	}
}
vector<vector<int> > combinationSum2(vector<int> &num, int target){
	_num = num;
	sort(_num.begin(), _num.end());
	vector<int> path;
	dfs(0, target, path);
	return res;
}

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