Leetcode dfs Combination Sum

Combination Sum

Total Accepted: 17319 Total Submissions: 65259My Submissions

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

The same repeated number may be chosen from C unlimited number of times.

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 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]

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中的同一数字能够拿多次。找到的组合不能反复。

思路：dfs
每一层的第i个节点有 n - i 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)

感觉測试数据有问题，我用以下两个代码。对于[1,1],1这个输入。输出的结果各自是[[1],[1]]和[[1]]，但两个代码都 Accepted 了。我感觉第二个代码才是正确的，输出结果没反复。

//代码一
vector<vector<int> > res;
vector<int> _nums;
void dfs(int target, int start, vector<int> &path){
	if(target == 0) res.push_back(path);
	for(int i = start; i < _nums.size(); ++i){
		if(target < _nums[i]) return ; //这里假设没剪枝的话会超时
		path.push_back(_nums[i]);
		dfs(target - _nums[i], i, path);
		path.pop_back();
	}
}


vector<vector<int> >combinationSum(vector<int> &nums, int target){
	_nums = nums;
	sort(_nums.begin(), _nums.end());
	vector<int> path;
	dfs(target, 0, path);
	return res;
}

//代码二
vector<vector<int> > res;
vector<int> _nums;
void dfs(int target, int start, vector<int> &path){
	if(target == 0) res.push_back(path);
	int previous = -1;
	for(int i = start; i < _nums.size(); ++i){
		if(_nums[i] == previous) continue;
		if(target < _nums[i]) return ; //这里假设没剪枝的话会超时
		previous = _nums[i];
		path.push_back(_nums[i]);
		dfs(target - _nums[i], i, path);
		path.pop_back();
	}
}

vector<vector<int> >combinationSum(vector<int> &nums, int target){
	_nums = nums;
	sort(_nums.begin(), _nums.end());
	vector<int> path;
	dfs(target, 0, path);
	return res;
}