刷题39. Combination Sum

一、题目说明

题目39. Combination Sum，是从正数列表中选取几个，其和等于目标数的可能组合。任何一个数可以重复取，如candidates = [2,3,6,7], target = 7,结果集合是[ [7], [2,2,3] ]

如candidates = [2,3,5], target = 8,结果集合是 [ [2,2,2,2], [2,3,3], [3,5] ]

题目难度是Medium，先思考一下，再来解答。

二、我的解答

经过一番思考，这个题目可以通过dfs（树的深度优先遍历）求解，首先我们画一个“树”，反映求解过程。这个图，我就不上了。我的代码：

#include
#include
#include
using namespace std;
class Solution{
    public:
        void dfs(vector>& res,vector& candidates,vector& path,int begin,int target){
            if(target==0){
                res.push_back(path);
                return;
            }
            for(int i=begin;i=0;i++){
                path.push_back(candidates[i]);
                dfs(res,candidates,path,i,target-candidates[i]);
                path.pop_back();
            }
        }
        vector> combinationSum(vector& candidates, int target){
            vector> res;
            vector path;
            if(candidates.size()<1){
                return res;
            }
            sort(candidates.begin(),candidates.end());
            dfs(res,candidates,path,0,target);
            return res;
        }
};
int main(){
    Solution s;
    vector candidates = {2,3,6,7};
    vector> res = s.combinationSum(candidates,7);
    cout<<"candidates of {2,3,6,7}"<<"\n";
    for(int i=0;i r = res[i];
        for(int j=0;jr = res[i];
        for(int j=0;j

性能：

Runtime: 12 ms, faster than 84.44% of C++ online submissions for Combination Sum.
Memory Usage: 9.8 MB, less than 58.33% of C++ online submissions for Combination Sum.

一行代码没修改，再次运行：

Runtime: 4 ms, faster than 99.94% of C++ online submissions for Combination Sum.
Memory Usage: 9.3 MB, less than 94.44% of C++ online submissions for Combination Sum.

三、优化

比较搞笑的是，我一行代码没修改，再次提交性能居然大幅提高。厉害了！

另外的解答思路是DP，这个是网上的，不是我写的。

用unordered_map> dict;存储数的分解，比如求{2,3,5,7}和是8的：

dict[2] ={2}

dict[3] = {3}

dict[4] = {2,2}

dict[5] = dict[2] + dict[3] = {2,3}

dict[6] = {dict[2] + dict[2] + dict[2]},{dict[3]}

....

class Solution {
public:
    vector> combinationSum(vector &candidates, int target)
    {
        unordered_map>> dict;
        for (int i = 1; i <= target; i++)
            for (int it : candidates)
                if (i == it){
                    dict[i].push_back(vector{ it });
                }
                else if (i > it){
                    for (auto ivec : dict[i - it]) {
                        if (it < ivec[ivec.size() - 1]) {
                            continue;
                        }
                        ivec.push_back(it);

                        dict[i].push_back(ivec);
                    }
                }

        return dict[target];
    }
};