LeetCode 39. 组合总和（C++、python）

给定一个无重复元素的数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。

candidates 中的数字可以无限制重复被选取。

说明：

• 所有数字（包括 target）都是正整数。
• 解集不能包含重复的组合。 

示例 1:

输入: candidates = [2,3,6,7], target = 7,
所求解集为:
[
  [7],
  [2,2,3]
]

示例 2:

输入: candidates = [2,3,5], target = 8,
所求解集为:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

C++

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

python

class Solution:
    def DFS(self, candidates, res, tmp, target, k, su):
        if su==target:
            res.append(tmp.copy())
            return
        elif su>target:
            return
        for i in range(k,len(candidates)):
            su+=candidates[i]
            tmp.append(candidates[i])
            self.DFS(candidates,res,tmp,target,i,su)
            del tmp[-1]
            su-=candidates[i]
               
    def combinationSum(self, candidates, target):
        \"\"\"
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        \"\"\"
        res=[]
        tmp=[]
        su=0
        self.DFS(candidates,res,tmp,target,0,su)
        return res