[Leetcode 52] 39 Combination Sum

Problem:

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]

Analysis:

It's also a backtracking problem. One thing need to be noticed is that [1, 2, 1] and [1, 1, 2] are considered the same sum, so they can't be in the result. Further more, since final results are required to be sorted, [1, 2, 1] is even an invalid result. The solution to this problem is that make sure the given array sorted and bc won't access any element before the element has been accessed. That's why we add pos into the bc function.

Code:

class Solution {
public:
    vector<vector<int> > res;
    int tar;

    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        res.clear();
        if (candidates.size() == 0)
            return res;
            
        vector<int> path;
        tar = target;
        sort(candidates.begin(), candidates.end());
        bc(0, path, 0, candidates);
        
        return res;
    }
    
    
    void bc(int sum, vector<int> path, int pos, vector<int> &cand) {
        if (sum > tar)
            return ;
            
        if (sum == tar) {
            res.push_back(path);
        }
        
        for (int i=pos; i<cand.size(); i++) {
            path.push_back(cand[i]);
            bc(sum+cand[i], path, i, cand);
            path.pop_back();
        }
        
        return ;
    }
};