LeetCode: 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, ... , a_k) must be in non-descending order. (ie, a₁ <= a₂ <= ... <= a_k).
• 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] 

This problem could be solved as a DFS, to maintain the result always in a non-descending order, the search tree should be: the child node is always larger than the parent node

1

1

1, 2, 3, ...

DFS算法，假定已经选择了[a1, a1, .. ak], 那么问题转化为 从[ak ... an]中，选出一定的组合，使得和为target - (a1+a1 + .. ak)

This code pass the large test result within 80 ms

class Solution {
     void find_subset(vector<int>::iterator first,vector<int>::iterator last, 
                      vector<int> &comb, int target, vector<vector<int>> &ret){
         if(target == 0)
             ret.push_back(comb);
         if(target < 0)
             return;
         vector<int>::iterator it;
         for(it = first; it != last && *it <= target; it++){
             comb.push_back(*it);
             find_subset(it, last, comb, target - *it, ret);
             comb.pop_back();
         }
         
     }
public:
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int>> ret;
        unique(candidates.begin(), candidates.end());
        sort(candidates.begin(), candidates.end());
        vector<int> a;
        
        find_subset(candidates.begin(), candidates.end(), a, target, ret);
        return ret;
    }
};