39. Combination Sum(回溯)

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]
]

class Solution {
    private List<List<Integer>> res = new ArrayList<>();
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<Integer> temp = new ArrayList<Integer>();
        help(temp,candidates,0,0,target);
        return res;
    }
    private void help(List<Integer> temp,int[] nums,int index,int cursum,int target){
        if(cursum>target)
            return;
        if(cursum==target)
            res.add(new ArrayList<Integer>(temp));
        for(int i = index;i<nums.length;i++){
            temp.add(nums[i]);
            help(temp,nums,i,cursum+nums[i],target);
            temp.remove(temp.size()-1);
        }
    }
}

2019.3.12

class Solution {
public:
    vector<vector<int>> finalres ;
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        if(candidates.size()==0) return finalres;
        vector<int> curres;
        help(0,curres,candidates,0,target);
        return finalres;
        
    }
    void help(int cursum,vector<int>& curres,vector<int>& candidates,int index,int target){
        if(cursum==target)
            finalres.push_back(curres);
        if(cursum>target)
            return;
        for(int i = index;i<candidates.size();i++){
            curres.push_back(candidates[i]);
            help(cursum,curres,candidates,i,target-candidates[i]);
            curres.pop_back();

        }
    }
};