39. Combination Sum

39. Combination Sum

题目：

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]
]

代码如下：

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<int> store;
        vector<vector<int>> answer;
        combinationGet(candidates, target, store, answer, 0);
        return answer;
    }
    void combinationGet(vector<int>& candidates,
     int target, vector<int>& store, 
     vector<vector<int>>& answer, int start) {
        if (!target) {
            answer.push_back(store);
            return;
        }
        for (int i = start; i < candidates.size() && 
        target >= candidates[i]; i++) {
            store.push_back(candidates[i]);
            combinationGet(candidates, target - candidates[i], store, answer, i);
            store.pop_back();
        }
    }
};

解题思路：

• 结果中元素应保持non-descending order, 应先排序；
• 结果可以包含重复元素，递归中直接传 i 值就可以了。