leetcode 39. Combination Sum

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本博客探讨了在给定候选数集合的情况下，如何找出所有可能的组合，使得这些组合的元素之和等于指定的目标数。允许重复选择候选数，并确保组合中的元素有序且不重复。

1.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]

2.思路

从一堆数组中选取一些数，使得和等于target。与下一题不同的是一个数可以重复多次。

class Solution {
public:
    void comb(vector<int> candidates, int index, int target, vector<vector<int>> &res, vector<int> &path)
    {
    	if(0>target)return;
		if(0==target){res.push_back(path);return;}
		for(int i= index; i<candidates.size();i++)
		{
			path.push_back(candidates[i]);
			comb(candidates,i,target-candidates[i],res,path);
			path.pop_back();
		}
	}
	vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        sort(candidates.begin(),candidates.end());
		vector<vector<int>> res;
		vector<int> path;
		comb(candidates,0,target,res,path);
		return res;
    }
};