【leetcode】Combination Sum I && III

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:
All numbers (including target) will be positive integers.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
The solution set must not contain duplicate combinations.
For example, given candidate set 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]

 

class Solution {
public:
    void com(vector<int> &candidates,int start,int sum,int target,vector<int> &path,vector<vector<int> > &res)
    {
        if(sum>target)
            return ;
        if(sum==target)
        {
            res.push_back(path);
            return ;
        }

        int len=candidates.size();
        for(int i=start;i<len;i++)
        {
            path.push_back(candidates[i]);
            com(candidates,i,sum+candidates[i],target,path,res);
            path.pop_back();
        }
    }

    vector<vector<int> > combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(),candidates.end());
        vector<vector<int> > res;
        vector<int> path;
        com(candidates,0,0,target,path,res);
        return res;
    }
};

 

III：

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Ensure that numbers within the set are sorted in ascending order.

Example 1:

Input: k = 3, n = 7

Output:

[[1,2,4]]

Example 2:

Input: k = 3, n = 9

Output:

class Solution {
public:
    vector<vector<int>> combinationSum3(int k, int n) {
        vector<vector<int> >res;
        vector<int>item;
        dfs(k,n,1,item,res);
        return res;
    }

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

 

[[1,2,6], [1,3,5], [2,3,4]]