39. Combination Sum

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

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

Note:

• All numbers (including target) will be positive integers.
• The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

和为目标数的组合，数可以重复使用

C++：

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int>> res ;
        vector<int> temp ;
        backtracking(candidates,res,temp,target,0) ;
        return res ;
    }
    
    void backtracking(vector<int> candidates ,vector<vector<int>>& res,vector<int>& temp,int target , int start){
        if(target == 0){
            res.push_back(temp) ;
            return ;
        }
        for(int i = start ; i < candidates.size() ; i++){
            if (candidates[i] <= target){
                temp.push_back(candidates[i]) ;
                backtracking(candidates,res,temp,target-candidates[i],i) ;
                temp.pop_back() ;
            }
        }
    }
};