leetcode [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]

 ]

题目大意：

给一个不重复的数组，寻找数组中数字的排列组合，使得这个排列组合之和等于target。数组中的数字可以重复使用，找到数组中全部的排列组合。

 

解法：

我首先想到的是回溯法，首先对数组中的数进行排序，tmp数组记录当前排列组合，tmpSum记录当前排列组合之和，index记录遍历到数组哪一个数了。

C++：

class Solution {
private:
    void combinationCore(vector<int>&candidates,int index,int tmpSum,int target,vector<int>&tmp,vector<vector<int>>&res){
        if(tmpSum>target) return;
        if(tmpSum==target) {
            res.push_back(tmp);
            return;
        }
        for(int i=index;i<candidates.size();i++){
            tmpSum+=candidates[i];
            tmp.push_back(candidates[i]);
            combinationCore(candidates,i,tmpSum,target,tmp,res);
            tmpSum-=candidates[i];
            tmp.pop_back();
        }
    }

public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int>>res;
        vector<int>tmp;
        sort(candidates.begin(),candidates.end());
        combinationCore(candidates,0,0,target,tmp,res);

        return res;
    }
};

　　其实自己写的代码有点冗余，其中的tmpSum可以不需要，直接使用target减去值，再和零进行比较，递归调用的时候也是。可以参考下面的python代码。

 

Python：

class Solution(object):
    def combinationSum(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        res=[]
        candidates.sort()
        self.dfs(candidates,target,0,[],res)
        return res

    def dfs(self,candidates,target,index,path,res):
        if target<0:
            return
        if target==0:
            res.append(path)
            return
        for i in range(index,len(candidates)):
            self.dfs(candidates,target-candidates[i],i,path+[candidates[i]],res)

　写上面的代码发现python中的list进行合并可以采用[]+[]

　range（start,end）返回的列表并不包括end