combination sum(I, II, III, IV)

II 简单dfs

vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        
        vector<vector<int>> ans;
        vector<int> cur;
        sort(candidates.begin(), candidates.end()); // 排序后去重
        dfs(0,target,candidates,cur,ans);
        return ans;
    }
    void dfs(int level, int target, vector<int>& candidates,vector<int>& cur,vector<vector<int>> &ans)
    {
        if(target==0)
        {
            ans.push_back(cur);
            return;
        }
        if(target<0)return;
        for(int i=level;i<candidates.size();i++)
        {
                if (i > level && candidates[i] == candidates[i - 1])continue;// 去重
                cur.push_back(candidates[i]);
                dfs(i+1,target-candidates[i],candidates,cur,ans);
                cur.pop_back();
        }
            
    }

III 简单dfs递归，限制条件是k个数其和为n

    vector<vector<int>> combinationSum3(int k, int n) {
        vector<vector<int>> ans;
        vector<int> cur;
        dfs(k,n,1,cur,ans);
        return ans;
    }
    void dfs(int k,int n, int level, vector<int> &cur, vector<vector<int>> &ans)
    {
        if(k==0&&n==0)
        {
            ans.push_back(cur);
            return;
        }
        if(k==0)return;
        for(int i=level;i<=9;i++)
        {
            cur.push_back(i);
            dfs(k-1,n-i,i+1,cur,ans);
            cur.pop_back();
        }
    }

IV 简单dp，dfs超时，记忆化dfs应该可以

        dp[0]=1;
        for(int i=1;i<=target;i++)
        {
            for(int j=0;j<nums.size();j++)
            {
                if(nums[j]<=i)dp[i]+=dp[i-nums[j]];
            }
        }
        return dp[target];