C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

same repeated number may be chosen from C

Note:

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

​​[2, 3, 6, 7]​​ and target ​​7​​, 

A solution set is: 


[ [7], [2, 2, 3] ]



class Solution {
public:
  vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
    vector<int>tmp;
    dfs(candidates, target, tmp, 0);
    return s;
  }
private:

  void dfs(vector<int>& candidates, int target, vector<int>&tmp,int cur){
    if (target < 0) return;
    if (target==0){
      s.push_back(tmp);
      return;
    }
    for (int i = cur; i < candidates.size(); i++){
      tmp.push_back(candidates[i]);
      dfs(candidates, target-candidates[i], tmp,i);//i，因为一个数字可以出现多次
      tmp.pop_back();
    }
  }

  vector<vector<int> > s;
};


或

class Solution {
public:
  vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
    sort(candidates.begin(), candidates.end());
    vector<int>tmp;
    dfs(candidates, target, tmp, 0);
    return s;
  }
private:

  void dfs(vector<int>& candidates, int target, vector<int>&tmp,int cur){
    if (target < 0) return;
    if (target==0){
      s.push_back(tmp);
      return;
    }
    for (int i = cur; i < candidates.size(); i++){
      if (cur != i&&candidates[i] == candidates[i-1]) continue;
      tmp.push_back(candidates[i]);
      dfs(candidates, target-candidates[i], tmp,i);
      tmp.pop_back();
    }
  }

  vector<vector<int> > s;
};