39. 組合總數

題目連結：https://leetcode-cn.com/problems/combination-sum/

class Solution {
public:
    vector<vector<int>> res;
    vector<int> path;
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        dfs(candidates, target, 0, 0);
        return res;
    }

    void dfs(vector<int>& candidates, int target, int u, int sum)
    {
        if(sum == target) 
        {
            res.push_back(path);
            return ;
        }

        //每次都從當前位置開始列舉下一個數，這樣保證後面的數都是比前面大不會重複
        for(int i = u; i < candidates.size(); i++)
        {
            if(sum + candidates[i] > target) break;
            path.push_back(candidates[i]);
            //從i開始列舉，保證當前數不會比前一個數大
            dfs(candidates, target, i, sum + candidates[i]);
            path.pop_back();
        }
        
    }
};
 

40. 組合總數II

題目連結：https://leetcode-cn.com/problems/combination-sum-ii/

class Solution {
public:
    vector<vector<int>> res;
    vector<int> path;
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        dfs(candidates, target, 0, 0);
        return res;
    }

    void dfs(vector<int>& candidates, int target, int u, int sum)
    {
        if(sum == target) 
        {
            res.push_back(path);
            return ;
        }

      
        for(int i = u; i < candidates.size(); i++)
        {
            if(i > u && candidates[i] == candidates[i - 1]) continue;
            if(sum + candidates[i] > target) break;
            path.push_back(candidates[i]);
            //從i開始列舉，保證當前數不會比前一個數大
            dfs(candidates, target, i + 1, sum + candidates[i]);
            path.pop_back();
        }
        
    }
};
 

216. 組合總數III

題目連結: https://leetcode-cn.com/problems/combination-sum-iii/

class Solution {
public:
    vector<vector<int>> res;
    vector<int> path;
    
    vector<vector<int>> combinationSum3(int k, int n) {
        dfs(n, k, 1);
        return res;
    }
    void dfs(int target, int k, int u)
    {
        if(!target && !k)
        {
            res.push_back(path);
            return ;
        }
        for(int i = u; i <= 9; i++)
        {

            int tmp = target - i;
            if(tmp < 0) break;

            path.push_back(i);
            dfs(tmp, k - 1, i + 1);
            path.pop_back();
        }
    }
};