LeetCode Combination Sum

最新推荐文章于 2019-05-21 16:08:00 发布

coobj2008

最新推荐文章于 2019-05-21 16:08:00 发布

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本文介绍了一种寻找候选数集合中所有唯一组合的算法，这些组合的和等于目标数。算法允许同一候选数被重复选择，并确保解决方案集不包含重复组合。

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题目：

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]

class Solution {
public:
	vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
		cs = 0;
		res.clear();
		sort(candidates.begin(), candidates.end());
		dfs(0, candidates.size(), candidates, target);
		return res;
	}
private:
	int a[1000];
	int cs;
	vector<vector<int> > res;
	void dfs(int dep, int maxDep, vector<int> &candidates, int &target) {
		if (cs > target)
			return;
		if (cs == target) {
			if (cs == target) {
				vector<int> ans;
				for (int i = 0; i < dep; i++) {
					for (int j = 0; j < a[i]; j++)
						ans.push_back(candidates[i]);
				}
				res.push_back(ans);
			}
			return;

		}
		if (dep == maxDep)
			return;
		for (int i = 0; i <= target / candidates[dep]; i++) {
			a[dep] = i;
			cs += candidates[dep] * i;
			dfs(dep + 1, maxDep, candidates, target);
			cs -= candidates[dep] * i;
			a[dep] = 0;
		}
	}
};