[LeetCode]Combination Sum

题目描述：

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] 

解题思路：

回溯法，具体如下：

对于[2, 3, 6, 7]， target = 7:

1. 第一个数字选择2，target = 7 -2 = 5;

2. 第二个数字继续选择2(可以重复)，target = 5 - 2 = 3;

3. 第三个数字继续选择2， target = 1;

4. 没有元素比target小，回退到第三步，target = 3;

5. 第三个数字选择3，target = 0.

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int>> result;
        vector<int> elem;
        combinationSum(candidates, 0, target, result, elem);
        return result;
    }
private:
    void combinationSum(vector<int> &candidates, int index, int target, vector<vector<int>> &result, vector<int> &elem) {
        if (target == 0) {
            result.push_back(elem);
            return;
        }
        
        for (int i = index; i < candidates.size() && candidates[i] <= target; ++i) {
            if (i == 0 || candidates[i] != candidates[i -1]) {
                elem.push_back(candidates[i]);
                combinationSum(candidates, i, target - candidates[i], result, elem);
                elem.pop_back();
            }
        }
    }
};