[LeetCode]Combination Sum

Combination Sum

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] 

将上题Combination Sum II稍微修改就行了。

dfs(candidates,target,i+1,tmp,result_set);

改成

dfs(candidates,target,i,tmp,result_set);

表示第i位会重复。

class Solution {
public:
    void dfs(vector<int> candidates,int target,int start,vector<int> tmp,set<vector<int>>& result_set)
    {
        int sum=0;
        for(int i=0;i<tmp.size();i++)
        {
            sum+=tmp[i];
        }
        if(sum==target)
        {
            sort(tmp.begin(),tmp.end());
            result_set.insert(tmp);
            return;
        }
        else if(sum>target)
        {
            return;
        }
        for(int i=start;i<candidates.size();i++)
        {
            tmp.push_back(candidates[i]);
            dfs(candidates,target,i,tmp,result_set);
            tmp.pop_back();
        }
    }
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<int> tmp;
        set<vector<int>> result_set;
        vector<vector<int>> result;
        dfs(candidates,target,0,tmp,result_set);
        set<vector<int>>::iterator it;
        for(it=result_set.begin();it!=result_set.end();it++)
        {
            result.push_back(*it);
        }
        return result;
    }
};