39. Combination Sum

Given a set of candidate numbers (C) (without duplicates) 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.
• 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]
] 

要求你從一串array中找出符合某個目標加總的集合

解法的邏輯都是以DFS為主 有點類似Subset的作法，但是不同點在於要符合條件的元素才可以加入

。

不多說，來看code

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        
        vector<vector<int>> answerset;
        vector<int> combination;
        sort(candidates.begin(), candidates.end());
        
        combinationSumDFS(candidates, 0, target -0, combination, answerset);
        
        return answerset;
    }
    
    void combinationSumDFS(const vector<int>& candidates, int start, int target, vector<int> &combination, vector<vector<int>>              
                           &answerset)
    {
        // 小於0代表，再找下去也不可能會有符合的數字了
        if (target < 0)
        {
            return;
        }
        // 這個數字組合剛好符合
        else if (target == 0)
        {
            answerset.push_back(combination);
        }
        else
        {
            for (int i = start; i < candidates.size(); i++)
            {
                combination.push_back(candidates[i]);
                combinationSumDFS(candidates, i, target - candidates[i], combination, answerset);
                combination.pop_back();
            }
        }        
    }
};