leetcode 39 Combination Sum

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本文探讨了如何从一组候选数中找到所有可能的组合，使得这些组合的和等于给定的目标数。允许重复选择候选数，并且组合必须按非递减顺序排列，确保结果中不包含重复组合。

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]

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class Solution {
public:
	void findNum(vector<int> &candidates, vector<int> container, int target, int i, int sum, vector<vector<int> > &ans) {
		int n = candidates.size();

		if(sum == target) {
			ans.push_back(container);
			return;
		}

		if(i >= n || sum > target || sum+candidates[i]>target) {

			return;
		}

		int cnt = (target-sum)/candidates[i];

		for(int j = 0; j <= cnt; j++) {
			for(int k=0; k < j; k++) {
				container.push_back(candidates[i]);
			}

			findNum(candidates, container, target, i+1, sum+candidates[i]*j, ans);

			for(int k = 0; k < j; k++) {
				container.erase(container.end()-1);
			}

		}

	}
	vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
		vector<int> container;
		vector<vector<int> > ans;

		sort(candidates.begin(), candidates.end());

		findNum(candidates, container, target, 0, 0, ans);

		reverse(ans.begin(), ans.end());

		return ans;
	}
}