[leetcode 39] Combination Sum

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本文介绍了一种算法，用于在给定的候选数集中找到所有可能的组合，使得这些组合的元素之和等于指定的目标数。算法允许重复选择候选数，并确保组合中的数按非降序排列，避免了重复解决方案。

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:
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
        vector<int> cur;
        vector<vector<int> > res;
        sort(candidates.begin(), candidates.end());
        dfs(candidates, 0, target, cur, res);
        return res;
        
    }
    void dfs(vector<int> &candidate, int start, int gap, vector<int> &cur, vector<vector<int> > &res) {
        if (start == candidate.size()) {
            return ;
        }
        if (gap == 0) {
            res.push_back(cur);
            return ;
        }
        
        for (int i = start; i < candidate.size(); i++) {
            if (gap < candidate[i]) {
                return ;
            }
            cur.push_back(candidate[i]);
            dfs(candidate, i, gap-candidate[i], cur, res);
            cur.pop_back();
        }
        
    }
};