leetcode-39. Combination Sum 组合总和

题目：

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] ]

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] ]

思路：看到这种求所有解的情况，类似于排列组合，可以用dfs或者bfs。此题用dfs写一个递归程序就可以求所有 可能情况了。

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

class Solution:
    def combinationSum(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        def dfs(candidates,target,start,path,res):
            if target==0:
                return res.append(path+[])
            for i in range(start,len(candidates)):
                if target-candidates[i]>=0:
                    path.append(candidates[i])
                    dfs(candidates,target-candidates[i],i,path,res)
                    path.pop()
        
        res=[]
        dfs(candidates,target,0,[],res)
        return res