LeetCode #39

# Description:

Given a set of candidate numbers (C) (without duplicates) 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. The solution set must not contain duplicate combinations.

## Example:

``````For example, given candidate set [2, 3, 6, 7] and target 7,
A solution set is:
[
[7],
[2, 2, 3]
]
``````

# Code:

``````class Solution {
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<vector<int>> result;
vector<int> curr_sequence;

sort(candidates.begin(), candidates.end(), [](int a, int b){return a>b;});

DFS(candidates, 0, target, curr_sequence, result);

return result;
}

void DFS(const vector<int>& nums, int loc, int remain, vector<int>& curr_sequence, vector<vector<int>>& result){
if(remain==0){
result.push_back(curr_sequence);
return;
}

for(int i=loc; i<nums.size(); ++i){
if(remain>=nums[i]){
curr_sequence.push_back(nums[i]);
DFS(nums, i, remain-nums[i], curr_sequence, result);
curr_sequence.pop_back();
}
}
}
};
``````